CBSE Class 10 · Maths
CBSE Class 10 Maths Introduction to Trigonometry Previous Year Questions
Help your child master CBSE Class 10 Maths Introduction to Trigonometry Previous Year Questions with this curated collection sourced from real board papers spanning 2019–2023. Every question comes with a detailed step-by-step solution, helping your child confidently tackle trigonometric identities, standard angle values, and proof-based questions — topics that consistently carry marks in the board exam.
CBSE Class 10 Maths Introduction to Trigonometry — Questions with Solutions
$\sec \theta$ when expressed in terms of $\cot \theta$ is equal to
Solution
Answer: Option (C) is correct.
Explanation: We know that $\sec^{2}\theta = 1 + \tan^{2}\theta$ and $\tan\theta = \dfrac{1}{\cot\theta}$.
Therefore: $$\sec^{2}\theta = 1 + \left(\frac{1}{\cot\theta}\right)^{2} = \frac{1 + \cot^{2}\theta}{\cot^{2}\theta}$$ Thus, $\sec\theta = \dfrac{\sqrt{1+\cot^{2}\theta}}{\cot\theta}$, which corresponds to Option (C).
Explanation: We know that $\sec^{2}\theta = 1 + \tan^{2}\theta$ and $\tan\theta = \dfrac{1}{\cot\theta}$.
Therefore: $$\sec^{2}\theta = 1 + \left(\frac{1}{\cot\theta}\right)^{2} = \frac{1 + \cot^{2}\theta}{\cot^{2}\theta}$$ Thus, $\sec\theta = \dfrac{\sqrt{1+\cot^{2}\theta}}{\cot\theta}$, which corresponds to Option (C).
Which of the following is true for all values of $\theta$ $\left(0^{\circ} \leq \theta \leq 90^{\circ}\right)$?
Solution
Answer: Option (C) is correct.
Explanation: From the standard identity, $\sec^{2}\theta = 1 + \tan^{2}\theta$, which gives $\sec^{2}\theta – \tan^{2}\theta = 1$. This identity holds for all values of $\theta$ in $\left[0^{\circ}, 90^{\circ}\right]$, corresponding to Option (C).
Explanation: From the standard identity, $\sec^{2}\theta = 1 + \tan^{2}\theta$, which gives $\sec^{2}\theta – \tan^{2}\theta = 1$. This identity holds for all values of $\theta$ in $\left[0^{\circ}, 90^{\circ}\right]$, corresponding to Option (C).
$\left(\sec^{2}\theta – 1\right)\left(\operatorname{cosec}^{2}\theta – 1\right)$ is equal to:
Solution
Answer: Option (B) is correct.
Explanation: Using the identities $\sec^{2}\theta – 1 = \tan^{2}\theta$ and $\operatorname{cosec}^{2}\theta – 1 = \cot^{2}\theta$: $$\left(\sec^{2}\theta – 1\right)\left(\operatorname{cosec}^{2}\theta – 1\right) = \tan^{2}\theta \times \cot^{2}\theta = \tan^{2}\theta \times \frac{1}{\tan^{2}\theta} = 1$$ This corresponds to Option (B).
Explanation: Using the identities $\sec^{2}\theta – 1 = \tan^{2}\theta$ and $\operatorname{cosec}^{2}\theta – 1 = \cot^{2}\theta$: $$\left(\sec^{2}\theta – 1\right)\left(\operatorname{cosec}^{2}\theta – 1\right) = \tan^{2}\theta \times \cot^{2}\theta = \tan^{2}\theta \times \frac{1}{\tan^{2}\theta} = 1$$ This corresponds to Option (B).
$$\frac{\cos^{2}\theta}{\sin^{2}\theta} – \frac{1}{\sin^{2}\theta}$$ in simplified form is?
Solution
Answer: Option (D) is correct.
Explanation: $$\frac{\cos^{2}\theta}{\sin^{2}\theta} – \frac{1}{\sin^{2}\theta} = \frac{\cos^{2}\theta – 1}{\sin^{2}\theta}$$ Since $\sin^{2}\theta + \cos^{2}\theta = 1$, we get $\cos^{2}\theta – 1 = -\sin^{2}\theta$. Substituting: $$\frac{-\sin^{2}\theta}{\sin^{2}\theta} = -1$$ This corresponds to Option (D).
Explanation: $$\frac{\cos^{2}\theta}{\sin^{2}\theta} – \frac{1}{\sin^{2}\theta} = \frac{\cos^{2}\theta – 1}{\sin^{2}\theta}$$ Since $\sin^{2}\theta + \cos^{2}\theta = 1$, we get $\cos^{2}\theta – 1 = -\sin^{2}\theta$. Substituting: $$\frac{-\sin^{2}\theta}{\sin^{2}\theta} = -1$$ This corresponds to Option (D).
If $\theta$ is an acute angle of a right angled triangle, then which of the following equations is NOT true?
Solution
Answer: Option (D) is correct.
Explanation: From the standard identity, $\tan^{2}\theta + 1 = \sec^{2}\theta$, which gives $\tan^{2}\theta – \sec^{2}\theta = -1$, not $+1$. Hence option (D) is false, corresponding to Option (D).
Explanation: From the standard identity, $\tan^{2}\theta + 1 = \sec^{2}\theta$, which gives $\tan^{2}\theta – \sec^{2}\theta = -1$, not $+1$. Hence option (D) is false, corresponding to Option (D).
$\left(\cos^{4}A – \sin^{4}A\right)$ on simplification gives:
Solution
Answer: Option (D) is correct.
Explanation: Using the difference of squares identity $a^{2} – b^{2} = (a+b)(a-b)$: $$\cos^{4}A – \sin^{4}A = \left(\cos^{2}A + \sin^{2}A\right)\left(\cos^{2}A – \sin^{2}A\right)$$ Since $\cos^{2}A + \sin^{2}A = 1$: $$= \cos^{2}A – \sin^{2}A = \cos^{2}A – (1 – \cos^{2}A) = 2\cos^{2}A – 1$$ This corresponds to Option (D).
Explanation: Using the difference of squares identity $a^{2} – b^{2} = (a+b)(a-b)$: $$\cos^{4}A – \sin^{4}A = \left(\cos^{2}A + \sin^{2}A\right)\left(\cos^{2}A – \sin^{2}A\right)$$ Since $\cos^{2}A + \sin^{2}A = 1$: $$= \cos^{2}A – \sin^{2}A = \cos^{2}A – (1 – \cos^{2}A) = 2\cos^{2}A – 1$$ This corresponds to Option (D).
If $\tan\theta = \dfrac{x}{y}$, then $\cos\theta$ is equal to:
Solution
Answer: Option (B) is correct.
Explanation: Since $\tan\theta = \dfrac{x}{y} = \dfrac{\text{Perpendicular}}{\text{Base}}$, we have $P = x$ and $B = y$.
By Pythagoras theorem: $H = \sqrt{x^{2} + y^{2}}$
Therefore: $\cos\theta = \dfrac{\text{Base}}{\text{Hypotenuse}} = \dfrac{y}{\sqrt{x^{2}+y^{2}}}$, which corresponds to Option (B).
Explanation: Since $\tan\theta = \dfrac{x}{y} = \dfrac{\text{Perpendicular}}{\text{Base}}$, we have $P = x$ and $B = y$.
By Pythagoras theorem: $H = \sqrt{x^{2} + y^{2}}$
Therefore: $\cos\theta = \dfrac{\text{Base}}{\text{Hypotenuse}} = \dfrac{y}{\sqrt{x^{2}+y^{2}}}$, which corresponds to Option (B).
If $4\cot\theta – 5 = 0$, then the value of $\dfrac{5\sin\theta – 4\cos\theta}{5\sin\theta + 4\cos\theta}$ is:
Solution
Answer: Option (C) is correct.
Explanation: Given $4\cot\theta – 5 = 0 \Rightarrow \cot\theta = \dfrac{5}{4}$, so $B = 5$, $P = 4$.
By Pythagoras theorem: $H = \sqrt{16 + 25} = \sqrt{41}$
Therefore $\sin\theta = \dfrac{4}{\sqrt{41}}$ and $\cos\theta = \dfrac{5}{\sqrt{41}}$. Substituting: $$\frac{5 \times \frac{4}{\sqrt{41}} – 4 \times \frac{5}{\sqrt{41}}}{5 \times \frac{4}{\sqrt{41}} + 4 \times \frac{5}{\sqrt{41}}} = \frac{\frac{20}{\sqrt{41}} – \frac{20}{\sqrt{41}}}{\frac{20}{\sqrt{41}} + \frac{20}{\sqrt{41}}} = \frac{0}{\frac{40}{\sqrt{41}}} = 0$$ This corresponds to Option (C).
Explanation: Given $4\cot\theta – 5 = 0 \Rightarrow \cot\theta = \dfrac{5}{4}$, so $B = 5$, $P = 4$.
By Pythagoras theorem: $H = \sqrt{16 + 25} = \sqrt{41}$
Therefore $\sin\theta = \dfrac{4}{\sqrt{41}}$ and $\cos\theta = \dfrac{5}{\sqrt{41}}$. Substituting: $$\frac{5 \times \frac{4}{\sqrt{41}} – 4 \times \frac{5}{\sqrt{41}}}{5 \times \frac{4}{\sqrt{41}} + 4 \times \frac{5}{\sqrt{41}}} = \frac{\frac{20}{\sqrt{41}} – \frac{20}{\sqrt{41}}}{\frac{20}{\sqrt{41}} + \frac{20}{\sqrt{41}}} = \frac{0}{\frac{40}{\sqrt{41}}} = 0$$ This corresponds to Option (C).
Given that $\sec\theta = \sqrt{2}$, the value of $\dfrac{1 + \tan\theta}{\sin\theta}$ is:
Solution
Answer: Option (A) is correct.
Explanation: Since $\sec\theta = \sqrt{2}$ and $\sec 45^{\circ} = \sqrt{2}$, we get $\theta = 45^{\circ}$.
Substituting: $$\frac{1 + \tan 45^{\circ}}{\sin 45^{\circ}} = \frac{1 + 1}{\frac{1}{\sqrt{2}}} = 2\sqrt{2}$$ This corresponds to Option (A).
Explanation: Since $\sec\theta = \sqrt{2}$ and $\sec 45^{\circ} = \sqrt{2}$, we get $\theta = 45^{\circ}$.
Substituting: $$\frac{1 + \tan 45^{\circ}}{\sin 45^{\circ}} = \frac{1 + 1}{\frac{1}{\sqrt{2}}} = 2\sqrt{2}$$ This corresponds to Option (A).
If $x\tan 60^{\circ}\cos 60^{\circ} = \sin 60^{\circ}\cot 60^{\circ}$, then $x =$
Solution
Answer: Option (B) is correct.
Explanation: $$x \times \sqrt{3} \times \frac{1}{2} = \frac{\sqrt{3}}{2} \times \frac{1}{\sqrt{3}}$$ $$x\sqrt{3} = 1 \Rightarrow x = \frac{1}{\sqrt{3}} = \tan 30^{\circ}$$ This corresponds to Option (B).
Explanation: $$x \times \sqrt{3} \times \frac{1}{2} = \frac{\sqrt{3}}{2} \times \frac{1}{\sqrt{3}}$$ $$x\sqrt{3} = 1 \Rightarrow x = \frac{1}{\sqrt{3}} = \tan 30^{\circ}$$ This corresponds to Option (B).
If $\theta$ is an acute angle and $\tan\theta + \cot\theta = 2$, then the value of $\sin^{3}\theta + \cos^{3}\theta$ is:
Solution
Answer: Option (C) is correct.
Explanation: From $\tan\theta + \cot\theta = 2$: $$\frac{\sin^{2}\theta + \cos^{2}\theta}{\cos\theta\sin\theta} = 2 \Rightarrow \sin\theta\cos\theta = \frac{1}{2} \quad \ldots(i)$$ Then: $(\sin\theta + \cos\theta)^{2} = 1 + 2\sin\theta\cos\theta = 1 + 1 = 2$, so $\sin\theta + \cos\theta = \sqrt{2}$ …(ii)
Using the identity $a^{3} + b^{3} = (a+b)^{3} – 3ab(a+b)$: $$\sin^{3}\theta + \cos^{3}\theta = (\sqrt{2})^{3} – 3 \times \frac{1}{2} \times \sqrt{2} = 2\sqrt{2} – \frac{3\sqrt{2}}{2} = \frac{\sqrt{2}}{2}$$ This corresponds to Option (C).
Explanation: From $\tan\theta + \cot\theta = 2$: $$\frac{\sin^{2}\theta + \cos^{2}\theta}{\cos\theta\sin\theta} = 2 \Rightarrow \sin\theta\cos\theta = \frac{1}{2} \quad \ldots(i)$$ Then: $(\sin\theta + \cos\theta)^{2} = 1 + 2\sin\theta\cos\theta = 1 + 1 = 2$, so $\sin\theta + \cos\theta = \sqrt{2}$ …(ii)
Using the identity $a^{3} + b^{3} = (a+b)^{3} – 3ab(a+b)$: $$\sin^{3}\theta + \cos^{3}\theta = (\sqrt{2})^{3} – 3 \times \frac{1}{2} \times \sqrt{2} = 2\sqrt{2} – \frac{3\sqrt{2}}{2} = \frac{\sqrt{2}}{2}$$ This corresponds to Option (C).
If $a\cot\theta + b\operatorname{cosec}\theta = p$ and $b\cot\theta + a\operatorname{cosec}\theta = q$, then $p^{2} – q^{2} =$
Solution
Answer: Option (B) is correct.
Explanation: $$p^{2} – q^{2} = (a\cot\theta + b\operatorname{cosec}\theta)^{2} – (b\cot\theta + a\operatorname{cosec}\theta)^{2}$$ Expanding and simplifying: $$= a^{2}(\cot^{2}\theta – \operatorname{cosec}^{2}\theta) + b^{2}(\operatorname{cosec}^{2}\theta – \cot^{2}\theta)$$ Since $\operatorname{cosec}^{2}\theta – \cot^{2}\theta = 1$: $$= a^{2}(-1) + b^{2}(1) = b^{2} – a^{2}$$ This corresponds to Option (B).
Explanation: $$p^{2} – q^{2} = (a\cot\theta + b\operatorname{cosec}\theta)^{2} – (b\cot\theta + a\operatorname{cosec}\theta)^{2}$$ Expanding and simplifying: $$= a^{2}(\cot^{2}\theta – \operatorname{cosec}^{2}\theta) + b^{2}(\operatorname{cosec}^{2}\theta – \cot^{2}\theta)$$ Since $\operatorname{cosec}^{2}\theta – \cot^{2}\theta = 1$: $$= a^{2}(-1) + b^{2}(1) = b^{2} – a^{2}$$ This corresponds to Option (B).
If $\sec\theta + \tan\theta = p$, then $\tan\theta$ is:
Solution
Answer: Option (B) is correct.
Explanation: Given $\sec\theta + \tan\theta = p$. Since $\sec\theta = \sqrt{1 + \tan^{2}\theta}$, substituting and squaring: $$1 + \tan^{2}\theta = p^{2} + \tan^{2}\theta – 2p\tan\theta$$ $$1 = p^{2} – 2p\tan\theta \Rightarrow \tan\theta = \frac{p^{2}-1}{2p}$$ This corresponds to Option (B).
Explanation: Given $\sec\theta + \tan\theta = p$. Since $\sec\theta = \sqrt{1 + \tan^{2}\theta}$, substituting and squaring: $$1 + \tan^{2}\theta = p^{2} + \tan^{2}\theta – 2p\tan\theta$$ $$1 = p^{2} – 2p\tan\theta \Rightarrow \tan\theta = \frac{p^{2}-1}{2p}$$ This corresponds to Option (B).
The value of $\dfrac{1}{\tan\theta + \cot\theta}$ is:
Solution
Answer: Option (A) is correct.
Explanation: $$\frac{1}{\tan\theta + \cot\theta} = \frac{1}{\frac{\sin\theta}{\cos\theta} + \frac{\cos\theta}{\sin\theta}} = \frac{1}{\frac{\sin^{2}\theta + \cos^{2}\theta}{\cos\theta\sin\theta}} = \frac{\cos\theta\sin\theta}{1} = \cos\theta\sin\theta$$ This corresponds to Option (A).
Explanation: $$\frac{1}{\tan\theta + \cot\theta} = \frac{1}{\frac{\sin\theta}{\cos\theta} + \frac{\cos\theta}{\sin\theta}} = \frac{1}{\frac{\sin^{2}\theta + \cos^{2}\theta}{\cos\theta\sin\theta}} = \frac{\cos\theta\sin\theta}{1} = \cos\theta\sin\theta$$ This corresponds to Option (A).
Given that $\sin\theta = \dfrac{a}{b}$, find $\cos\theta$.
Solution
Answer: Option (C) is correct.
Explanation: Given $\sin\theta = \dfrac{BC}{AC} = \dfrac{a}{b}$, so $BC = a$ and $AC = b$.
By Pythagoras theorem: $AC^{2} = AB^{2} + BC^{2}$, so $b^{2} = AB^{2} + a^{2}$, giving $AB = \sqrt{b^{2} – a^{2}}$.
Therefore: $\cos\theta = \dfrac{AB}{AC} = \dfrac{\sqrt{b^{2}-a^{2}}}{b}$, which corresponds to Option (C).
Explanation: Given $\sin\theta = \dfrac{BC}{AC} = \dfrac{a}{b}$, so $BC = a$ and $AC = b$.
By Pythagoras theorem: $AC^{2} = AB^{2} + BC^{2}$, so $b^{2} = AB^{2} + a^{2}$, giving $AB = \sqrt{b^{2} – a^{2}}$.
Therefore: $\cos\theta = \dfrac{AB}{AC} = \dfrac{\sqrt{b^{2}-a^{2}}}{b}$, which corresponds to Option (C).
Prove the following: $\dfrac{1}{\operatorname{cosec}\theta – \cot\theta} – \dfrac{\cot\theta}{\cos\theta} = \cot\theta$
Answer
Rationalising the first term by multiplying numerator and denominator by $(\operatorname{cosec}\theta + \cot\theta)$:
$$\frac{1}{\operatorname{cosec}\theta – \cot\theta} \times \frac{\operatorname{cosec}\theta + \cot\theta}{\operatorname{cosec}\theta + \cot\theta} – \frac{\cos\theta}{\sin\theta} \times \frac{1}{\cos\theta}$$ $$= \frac{\operatorname{cosec}\theta + \cot\theta}{\operatorname{cosec}^{2}\theta – \cot^{2}\theta} – \frac{1}{\sin\theta}$$ Since $\operatorname{cosec}^{2}\theta – \cot^{2}\theta = 1$: $$= \operatorname{cosec}\theta + \cot\theta – \operatorname{cosec}\theta = \cot\theta = \text{RHS}$$ Hence Proved.
$$\frac{1}{\operatorname{cosec}\theta – \cot\theta} \times \frac{\operatorname{cosec}\theta + \cot\theta}{\operatorname{cosec}\theta + \cot\theta} – \frac{\cos\theta}{\sin\theta} \times \frac{1}{\cos\theta}$$ $$= \frac{\operatorname{cosec}\theta + \cot\theta}{\operatorname{cosec}^{2}\theta – \cot^{2}\theta} – \frac{1}{\sin\theta}$$ Since $\operatorname{cosec}^{2}\theta – \cot^{2}\theta = 1$: $$= \operatorname{cosec}\theta + \cot\theta – \operatorname{cosec}\theta = \cot\theta = \text{RHS}$$ Hence Proved.
If $x = 2\sin^{2}\theta$ and $y = 2\cos^{2}\theta + 1$, then find the value of $x + y$.
Answer
$$x + y = 2\sin^{2}\theta + 2\cos^{2}\theta + 1 = 2(\sin^{2}\theta + \cos^{2}\theta) + 1$$
Since $\sin^{2}\theta + \cos^{2}\theta = 1$:
$$= 2 \times 1 + 1 = \mathbf{3}$$
Find the value of $\left(\sin^{2}\theta + \dfrac{1}{1 + \tan^{2}\theta}\right)$.
Answer
Since $1 + \tan^{2}\theta = \sec^{2}\theta$:
$$\sin^{2}\theta + \frac{1}{\sec^{2}\theta} = \sin^{2}\theta + \cos^{2}\theta = \mathbf{1}$$
Evaluate: $\dfrac{5\cos^{2}60^{\circ} + 4\sec^{2}30^{\circ} – \tan^{2}45^{\circ}}{\sin^{2}30^{\circ} + \cos^{2}30^{\circ}}$
Answer
Substituting standard values $\cos 60^{\circ} = \dfrac{1}{2}$, $\sec 30^{\circ} = \dfrac{2}{\sqrt{3}}$, $\tan 45^{\circ} = 1$, and using $\sin^{2}30^{\circ} + \cos^{2}30^{\circ} = 1$:
$$= \frac{5\left(\frac{1}{2}\right)^{2} + 4\left(\frac{2}{\sqrt{3}}\right)^{2} – 1}{1} = \frac{\frac{5}{4} + \frac{16}{3} – 1}{1}$$
$$= \frac{15 + 64 – 12}{12} = \boxed{\dfrac{67}{12}}$$
Prove that: $1 + \dfrac{\cot^{2}\alpha}{1 + \operatorname{cosec}\alpha} = \operatorname{cosec}\alpha$
Answer
$$\text{LHS} = 1 + \frac{\cot^{2}\alpha}{1 + \operatorname{cosec}\alpha} = 1 + \frac{\operatorname{cosec}^{2}\alpha – 1}{1 + \operatorname{cosec}\alpha}$$
Factorising the numerator using $a^{2} – b^{2} = (a+b)(a-b)$:
$$= 1 + \frac{(1 + \operatorname{cosec}\alpha)(\operatorname{cosec}\alpha – 1)}{1 + \operatorname{cosec}\alpha} = 1 + \operatorname{cosec}\alpha – 1 = \operatorname{cosec}\alpha = \text{RHS}$$ Hence Proved.
Show that $\tan^{4}\theta + \tan^{2}\theta = \sec^{4}\theta – \sec^{2}\theta$
Answer
$$\text{LHS} = \tan^{4}\theta + \tan^{2}\theta = \tan^{2}\theta(1 + \tan^{2}\theta) = \tan^{2}\theta \cdot \sec^{2}\theta$$
Since $\tan^{2}\theta = \sec^{2}\theta – 1$:
$$= (\sec^{2}\theta – 1)\sec^{2}\theta = \sec^{4}\theta – \sec^{2}\theta = \text{RHS}$$ Hence Proved.
Prove that $\dfrac{1 + \sec A}{\sec A} = \dfrac{\sin^{2}A}{1 – \cos A}$
Answer
LHS: $$\frac{1 + \sec A}{\sec A} = \frac{1 + \frac{1}{\cos A}}{\frac{1}{\cos A}} = \frac{\frac{\cos A + 1}{\cos A}}{\frac{1}{\cos A}} = \cos A + 1 \quad \ldots(i)$$ RHS: $$\frac{\sin^{2}A}{1 – \cos A} = \frac{1 – \cos^{2}A}{1 – \cos A} = \frac{(1 + \cos A)(1 – \cos A)}{1 – \cos A} = 1 + \cos A \quad \ldots(ii)$$
From (i) and (ii): LHS = RHS. Hence Proved.
If $1 + \sin^{2}\theta = 3\sin\theta\cos\theta$, then prove that $\tan\theta = 1$.
Answer
Dividing both sides by $\cos^{2}\theta$:
$$\sec^{2}\theta + \tan^{2}\theta = 3\tan\theta$$
$$1 + \tan^{2}\theta + \tan^{2}\theta = 3\tan\theta \quad (\because \sec^{2}\theta = 1 + \tan^{2}\theta)$$
$$2\tan^{2}\theta – 3\tan\theta + 1 = 0$$
Let $\tan\theta = x$: $(x – 1)(2x – 1) = 0 \Rightarrow x = 1$ or $x = \dfrac{1}{2}$
Therefore $\tan\theta = 1$ or $\dfrac{1}{2}$. Hence Proved (for $\tan\theta = 1$).
Therefore $\tan\theta = 1$ or $\dfrac{1}{2}$. Hence Proved (for $\tan\theta = 1$).
Prove that: $\dfrac{\sin A – 2\sin^{3}A}{2\cos^{3}A – \cos A} = \tan A$
Answer
$$\text{LHS} = \frac{\sin A(1 – 2\sin^{2}A)}{\cos A(2\cos^{2}A – 1)}$$
Substituting $\sin^{2}A = 1 – \cos^{2}A$ in the numerator:
$$= \frac{\sin A\left(1 – 2(1 – \cos^{2}A)\right)}{\cos A(2\cos^{2}A – 1)} = \frac{\sin A(2\cos^{2}A – 1)}{\cos A(2\cos^{2}A – 1)} = \frac{\sin A}{\cos A} = \tan A = \text{RHS}$$ Hence Proved.
Prove that: $\left(\sin^{4}\theta – \cos^{4}\theta + 1\right)\operatorname{cosec}^{2}\theta = 2$
Answer
$$\text{LHS} = \left(\sin^{4}\theta – \cos^{4}\theta + 1\right)\operatorname{cosec}^{2}\theta$$
Factorising using $a^{2} – b^{2} = (a+b)(a-b)$:
$$= \left[(\sin^{2}\theta – \cos^{2}\theta)(\sin^{2}\theta + \cos^{2}\theta) + 1\right]\operatorname{cosec}^{2}\theta$$
Since $\sin^{2}\theta + \cos^{2}\theta = 1$:
$$= (\sin^{2}\theta – \cos^{2}\theta + 1)\operatorname{cosec}^{2}\theta = 2\sin^{2}\theta \cdot \operatorname{cosec}^{2}\theta = 2 \times 1 = 2 = \text{RHS}$$ Hence Proved.
If $\sin x + \operatorname{cosec} x = 2$, then find the value of $\sin^{19}x + \operatorname{cosec}^{20}x$.
Answer
Given $\sin x + \dfrac{1}{\sin x} = 2$, multiplying through by $\sin x$:
$$\sin^{2}x + 1 = 2\sin x \Rightarrow (\sin x – 1)^{2} = 0 \Rightarrow \sin x = 1$$
Therefore $\operatorname{cosec} x = 1$.
$$\sin^{19}x + \operatorname{cosec}^{20}x = 1^{19} + 1^{20} = 1 + 1 = \mathbf{2}$$
$$\sin^{19}x + \operatorname{cosec}^{20}x = 1^{19} + 1^{20} = 1 + 1 = \mathbf{2}$$
If $\sin\theta + \cos\theta = p$ and $\sec\theta + \operatorname{cosec}\theta = q$, then prove that $q(p^{2} – 1) = 2p$.
Answer
LHS: $$q(p^{2}-1) = (\sec\theta + \operatorname{cosec}\theta)\left[(\sin\theta + \cos\theta)^{2} – 1\right]$$
$$= \left(\frac{1}{\cos\theta} + \frac{1}{\sin\theta}\right)\left(\sin^{2}\theta + \cos^{2}\theta + 2\sin\theta\cos\theta – 1\right)$$
Since $\sin^{2}\theta + \cos^{2}\theta = 1$:
$$= \frac{\sin\theta + \cos\theta}{\sin\theta\cos\theta} \times 2\sin\theta\cos\theta = 2(\sin\theta + \cos\theta) = 2p = \text{RHS}$$ Hence Proved.
$\sin\theta + \cos\theta = \sqrt{3}$, then prove that $\tan\theta + \cot\theta = 1$.
Answer
Squaring both sides: $\sin^{2}\theta + \cos^{2}\theta + 2\sin\theta\cos\theta = 3$
$\Rightarrow 1 + 2\sin\theta\cos\theta = 3 \Rightarrow \sin\theta\cos\theta = 1 \quad \ldots(i)$
LHS: $$\tan\theta + \cot\theta = \frac{\sin\theta}{\cos\theta} + \frac{\cos\theta}{\sin\theta} = \frac{\sin^{2}\theta + \cos^{2}\theta}{\cos\theta\sin\theta} = \frac{1}{1} = 1 = \text{RHS} \quad \text{[From (i)]}$$ Hence Proved.
$\Rightarrow 1 + 2\sin\theta\cos\theta = 3 \Rightarrow \sin\theta\cos\theta = 1 \quad \ldots(i)$
LHS: $$\tan\theta + \cot\theta = \frac{\sin\theta}{\cos\theta} + \frac{\cos\theta}{\sin\theta} = \frac{\sin^{2}\theta + \cos^{2}\theta}{\cos\theta\sin\theta} = \frac{1}{1} = 1 = \text{RHS} \quad \text{[From (i)]}$$ Hence Proved.
Prove that: $2\left(\sin^{6}\theta + \cos^{6}\theta\right) – 3\left(\sin^{4}\theta + \cos^{4}\theta\right) + 1 = 0$
Answer
Using the identity $a^{3} + b^{3} = (a+b)(a^{2} – ab + b^{2})$ with $a = \sin^{2}\theta$, $b = \cos^{2}\theta$:
$$\sin^{6}\theta + \cos^{6}\theta = (\sin^{2}\theta + \cos^{2}\theta)(\sin^{4}\theta – \sin^{2}\theta\cos^{2}\theta + \cos^{4}\theta) = \sin^{4}\theta – \sin^{2}\theta\cos^{2}\theta + \cos^{4}\theta$$
Substituting into LHS:
$$= 2(\sin^{4}\theta – \sin^{2}\theta\cos^{2}\theta + \cos^{4}\theta) – 3(\sin^{4}\theta + \cos^{4}\theta) + 1$$
$$= -\sin^{4}\theta – \cos^{4}\theta – 2\sin^{2}\theta\cos^{2}\theta + 1$$
$$= -(\sin^{2}\theta + \cos^{2}\theta)^{2} + 1 = -1 + 1 = 0 = \text{RHS}$$ Hence Proved.
Prove that: $\dfrac{\tan^{3}\theta}{1 + \tan^{2}\theta} + \dfrac{\cot^{3}\theta}{1 + \cot^{2}\theta} = \sec\theta\operatorname{cosec}\theta – 2\sin\theta\cos\theta$
Answer
Expressing in terms of $\sin\theta$ and $\cos\theta$:
$$\text{LHS} = \frac{\frac{\sin^{3}\theta}{\cos^{3}\theta}}{\frac{\sin^{2}\theta + \cos^{2}\theta}{\cos^{2}\theta}} + \frac{\frac{\cos^{3}\theta}{\sin^{3}\theta}}{\frac{\sin^{2}\theta + \cos^{2}\theta}{\sin^{2}\theta}} = \frac{\sin^{3}\theta}{\cos\theta} + \frac{\cos^{3}\theta}{\sin\theta}$$
$$= \frac{\sin^{4}\theta + \cos^{4}\theta}{\cos\theta\sin\theta} = \frac{(\sin^{2}\theta + \cos^{2}\theta)^{2} – 2\sin^{2}\theta\cos^{2}\theta}{\cos\theta\sin\theta}$$
$$= \frac{1 – 2\sin^{2}\theta\cos^{2}\theta}{\cos\theta\sin\theta} = \frac{1}{\cos\theta\sin\theta} – 2\sin\theta\cos\theta = \sec\theta\operatorname{cosec}\theta – 2\sin\theta\cos\theta = \text{RHS}$$ Hence Proved.
If $\dfrac{1}{\sin\theta – \cos\theta} = \dfrac{\operatorname{cosec}\theta}{\sqrt{2}}$, prove that $\left(\dfrac{1}{\sin\theta + \cos\theta}\right)^{2} = \dfrac{\sec^{2}\theta}{2}$.
Answer
Squaring the given condition:
$$\frac{1}{\sin^{2}\theta + \cos^{2}\theta – 2\sin\theta\cos\theta} = \frac{\operatorname{cosec}^{2}\theta}{2}$$
$$\frac{1}{1 – 2\sin\theta\cos\theta} = \frac{1}{2\sin^{2}\theta}$$
Cross-multiplying: $2\sin^{2}\theta = 1 – 2\sin\theta\cos\theta$, so $2\sin\theta\cos\theta = 1 – 2\sin^{2}\theta \quad \ldots(i)$
Now proving the required result: $$\text{LHS} = \frac{1}{\sin^{2}\theta + \cos^{2}\theta + 2\sin\theta\cos\theta} = \frac{1}{1 + 2\sin\theta\cos\theta}$$ Substituting from (i): $2\sin\theta\cos\theta = 1 – 2\sin^{2}\theta$: $$= \frac{1}{1 + (1 – 2\sin^{2}\theta)} = \frac{1}{2 – 2\sin^{2}\theta} = \frac{1}{2\cos^{2}\theta} = \frac{\sec^{2}\theta}{2} = \text{RHS}$$ Hence Proved.
Now proving the required result: $$\text{LHS} = \frac{1}{\sin^{2}\theta + \cos^{2}\theta + 2\sin\theta\cos\theta} = \frac{1}{1 + 2\sin\theta\cos\theta}$$ Substituting from (i): $2\sin\theta\cos\theta = 1 – 2\sin^{2}\theta$: $$= \frac{1}{1 + (1 – 2\sin^{2}\theta)} = \frac{1}{2 – 2\sin^{2}\theta} = \frac{1}{2\cos^{2}\theta} = \frac{\sec^{2}\theta}{2} = \text{RHS}$$ Hence Proved.
Prove that: $\dfrac{\sin A – \cos A + 1}{\sin A + \cos A – 1} = \dfrac{1}{\sec A – \tan A}$
Answer
Multiplying LHS numerator and denominator by $(1 + \sin A)$:
$$\text{LHS} = \frac{(\sin A – \cos A + 1)(1 + \sin A)}{(\sin A + \cos A – 1)(1 + \sin A)}$$
Simplifying the denominator: $\sin A + \cos A – 1 + \sin^{2}A + \cos A\sin A – \sin A$
$= \cos A – \cos^{2}A + \sin A\cos A = \cos A(1 – \cos A + \sin A)$
Therefore: $$= \frac{(1 + \sin A)}{\cos A} = \sec A + \tan A$$ Multiplying numerator and denominator by $(\sec A – \tan A)$: $$= \frac{(\sec A + \tan A)(\sec A – \tan A)}{\sec A – \tan A} = \frac{\sec^{2}A – \tan^{2}A}{\sec A – \tan A} = \frac{1}{\sec A – \tan A} = \text{RHS}$$ Hence Proved.
Therefore: $$= \frac{(1 + \sin A)}{\cos A} = \sec A + \tan A$$ Multiplying numerator and denominator by $(\sec A – \tan A)$: $$= \frac{(\sec A + \tan A)(\sec A – \tan A)}{\sec A – \tan A} = \frac{\sec^{2}A – \tan^{2}A}{\sec A – \tan A} = \frac{1}{\sec A – \tan A} = \text{RHS}$$ Hence Proved.
Prove that: $\dfrac{\sin\theta}{\cot\theta + \operatorname{cosec}\theta} = 2 + \dfrac{\sin\theta}{\cot\theta – \operatorname{cosec}\theta}$
Answer
Rearranging, it suffices to prove: $\dfrac{\sin\theta}{\operatorname{cosec}\theta + \cot\theta} – \dfrac{\sin\theta}{\cot\theta – \operatorname{cosec}\theta} = 2$
LHS becomes: $$\frac{\sin\theta}{\operatorname{cosec}\theta + \cot\theta} + \frac{\sin\theta}{\operatorname{cosec}\theta – \cot\theta} = \frac{\sin\theta\left[\operatorname{cosec}\theta – \cot\theta + \operatorname{cosec}\theta + \cot\theta\right]}{\operatorname{cosec}^{2}\theta – \cot^{2}\theta}$$ $$= \frac{\sin\theta(2\operatorname{cosec}\theta)}{1} = 2\sin\theta \cdot \frac{1}{\sin\theta} = 2 = \text{RHS}$$ Hence Proved.
LHS becomes: $$\frac{\sin\theta}{\operatorname{cosec}\theta + \cot\theta} + \frac{\sin\theta}{\operatorname{cosec}\theta – \cot\theta} = \frac{\sin\theta\left[\operatorname{cosec}\theta – \cot\theta + \operatorname{cosec}\theta + \cot\theta\right]}{\operatorname{cosec}^{2}\theta – \cot^{2}\theta}$$ $$= \frac{\sin\theta(2\operatorname{cosec}\theta)}{1} = 2\sin\theta \cdot \frac{1}{\sin\theta} = 2 = \text{RHS}$$ Hence Proved.
Prove that: $\dfrac{\tan^{2}A}{\tan^{2}A – 1} + \dfrac{\operatorname{cosec}^{2}A}{\sec^{2}A – \operatorname{cosec}^{2}A} = \dfrac{1}{1 – 2\cos^{2}A}$
Answer
Expressing in terms of $\sin A$ and $\cos A$:
$$\text{LHS} = \frac{\frac{\sin^{2}A}{\cos^{2}A}}{\frac{\sin^{2}A – \cos^{2}A}{\cos^{2}A}} + \frac{\frac{1}{\sin^{2}A}}{\frac{\sin^{2}A – \cos^{2}A}{\sin^{2}A\cos^{2}A}}$$
$$= \frac{\sin^{2}A}{\sin^{2}A – \cos^{2}A} + \frac{\cos^{2}A}{\sin^{2}A – \cos^{2}A} = \frac{\sin^{2}A + \cos^{2}A}{\sin^{2}A – \cos^{2}A}$$
$$= \frac{1}{\sin^{2}A – \cos^{2}A} = \frac{1}{(1 – \cos^{2}A) – \cos^{2}A} = \frac{1}{1 – 2\cos^{2}A} = \text{RHS}$$ Hence Proved.
Prove that: $(\tan\theta + \sec\theta – 1) \cdot (\tan\theta + 1 + \sec\theta) = \dfrac{2\sin\theta}{1 – \sin\theta}$
Answer
Let LHS $= (\tan\theta + \sec\theta – 1)(\tan\theta + \sec\theta + 1)$.
Treating $(\tan\theta + \sec\theta)$ as one term, this is of the form $(A – 1)(A + 1) = A^{2} – 1$: $$= (\tan\theta + \sec\theta)^{2} – 1 = \tan^{2}\theta + 2\tan\theta\sec\theta + \sec^{2}\theta – 1$$ Using $\sec^{2}\theta – 1 = \tan^{2}\theta$: $$= 2\tan^{2}\theta + 2\tan\theta\sec\theta = 2\tan\theta(\tan\theta + \sec\theta)$$ $$= 2 \cdot \frac{\sin\theta}{\cos\theta} \cdot \frac{1 + \sin\theta}{\cos\theta} \cdot \frac{\cos^{2}\theta}{1+\sin\theta}$$ Expressing directly: $$= \frac{2\sin\theta(1 + \sin\theta)}{\cos^{2}\theta} = \frac{2\sin\theta(1 + \sin\theta)}{1 – \sin^{2}\theta} = \frac{2\sin\theta(1 + \sin\theta)}{(1-\sin\theta)(1+\sin\theta)} = \frac{2\sin\theta}{1 – \sin\theta} = \text{RHS}$$ Hence Proved.
Treating $(\tan\theta + \sec\theta)$ as one term, this is of the form $(A – 1)(A + 1) = A^{2} – 1$: $$= (\tan\theta + \sec\theta)^{2} – 1 = \tan^{2}\theta + 2\tan\theta\sec\theta + \sec^{2}\theta – 1$$ Using $\sec^{2}\theta – 1 = \tan^{2}\theta$: $$= 2\tan^{2}\theta + 2\tan\theta\sec\theta = 2\tan\theta(\tan\theta + \sec\theta)$$ $$= 2 \cdot \frac{\sin\theta}{\cos\theta} \cdot \frac{1 + \sin\theta}{\cos\theta} \cdot \frac{\cos^{2}\theta}{1+\sin\theta}$$ Expressing directly: $$= \frac{2\sin\theta(1 + \sin\theta)}{\cos^{2}\theta} = \frac{2\sin\theta(1 + \sin\theta)}{1 – \sin^{2}\theta} = \frac{2\sin\theta(1 + \sin\theta)}{(1-\sin\theta)(1+\sin\theta)} = \frac{2\sin\theta}{1 – \sin\theta} = \text{RHS}$$ Hence Proved.
Case Study — Read the following and answer any four questions from Q.1 to Q.5.
In the figure below, PQRS is a quadrilateral. PR is perpendicular to QR and PS.
1. What is the value of $\tan Q$?
(A) $\dfrac{3}{5}$ (B) $\dfrac{1}{2}$ (C) $1$ (D) $\dfrac{4}{3}$
2. What is the length of RS?
(A) 8 units (B) 10 units (C) $8\sqrt{2}$ units (D) $\dfrac{16}{3}\sqrt{3}$ units
3. What is the value of $\sec\theta(1 – \sin\theta)(\sec\theta + \tan\theta)$?
(A) $1$ (B) $\sin^{2}\theta$ (C) $\sin\theta$ (D) $\cos\theta$
4. The length of PS is:
(A) $\sqrt{3}$ cm (B) $\dfrac{8}{3}$ cm (C) $\dfrac{8\sqrt{3}}{3}$ cm (D) $\dfrac{2\sqrt{3}}{3}$ cm
5. The value of $\tan 30^{\circ}\cot 60^{\circ}$ is:
(A) $1$ (B) $\dfrac{1}{3}$ (C) $\dfrac{1}{\sqrt{3}}$ (D) $\sqrt{3}$
In the figure below, PQRS is a quadrilateral. PR is perpendicular to QR and PS.
1. What is the value of $\tan Q$?
(A) $\dfrac{3}{5}$ (B) $\dfrac{1}{2}$ (C) $1$ (D) $\dfrac{4}{3}$
2. What is the length of RS?
(A) 8 units (B) 10 units (C) $8\sqrt{2}$ units (D) $\dfrac{16}{3}\sqrt{3}$ units
3. What is the value of $\sec\theta(1 – \sin\theta)(\sec\theta + \tan\theta)$?
(A) $1$ (B) $\sin^{2}\theta$ (C) $\sin\theta$ (D) $\cos\theta$
4. The length of PS is:
(A) $\sqrt{3}$ cm (B) $\dfrac{8}{3}$ cm (C) $\dfrac{8\sqrt{3}}{3}$ cm (D) $\dfrac{2\sqrt{3}}{3}$ cm
5. The value of $\tan 30^{\circ}\cot 60^{\circ}$ is:
(A) $1$ (B) $\dfrac{1}{3}$ (C) $\dfrac{1}{\sqrt{3}}$ (D) $\sqrt{3}$
Answer
1. Ans. Option (D) is correct.
In right angled $\triangle QPR$: $PR = \sqrt{PQ^{2} – QR^{2}} = \sqrt{10^{2} – 6^{2}} = \sqrt{64} = 8$ cm
$\therefore \tan Q = \dfrac{PR}{QR} = \dfrac{8}{6} = \dfrac{4}{3}$
2. Ans. Option (D) is correct.
In right angled $\triangle RPS$: $\sin 60^{\circ} = \dfrac{PR}{RS} \Rightarrow \dfrac{\sqrt{3}}{2} = \dfrac{8}{RS} \Rightarrow RS = \dfrac{16}{\sqrt{3}} = \dfrac{16\sqrt{3}}{3}$ units
3. Ans. Option (A) is correct.
$\sec\theta(1 – \sin\theta)(\sec\theta + \tan\theta) = \dfrac{1}{\cos\theta}(1-\sin\theta)\dfrac{(1+\sin\theta)}{\cos\theta} = \dfrac{1-\sin^{2}\theta}{\cos^{2}\theta} = \dfrac{\cos^{2}\theta}{\cos^{2}\theta} = 1$
4. Ans. Option (C) is correct.
$\tan 60^{\circ} = \dfrac{PR}{PS} \Rightarrow \sqrt{3} = \dfrac{8}{PS} \Rightarrow PS = \dfrac{8}{\sqrt{3}} = \dfrac{8\sqrt{3}}{3}$ cm
5. Ans. Option (B) is correct.
$\tan 30^{\circ}\cot 60^{\circ} = \dfrac{1}{\sqrt{3}} \times \dfrac{1}{\sqrt{3}} = \dfrac{1}{3}$
In right angled $\triangle QPR$: $PR = \sqrt{PQ^{2} – QR^{2}} = \sqrt{10^{2} – 6^{2}} = \sqrt{64} = 8$ cm
$\therefore \tan Q = \dfrac{PR}{QR} = \dfrac{8}{6} = \dfrac{4}{3}$
2. Ans. Option (D) is correct.
In right angled $\triangle RPS$: $\sin 60^{\circ} = \dfrac{PR}{RS} \Rightarrow \dfrac{\sqrt{3}}{2} = \dfrac{8}{RS} \Rightarrow RS = \dfrac{16}{\sqrt{3}} = \dfrac{16\sqrt{3}}{3}$ units
3. Ans. Option (A) is correct.
$\sec\theta(1 – \sin\theta)(\sec\theta + \tan\theta) = \dfrac{1}{\cos\theta}(1-\sin\theta)\dfrac{(1+\sin\theta)}{\cos\theta} = \dfrac{1-\sin^{2}\theta}{\cos^{2}\theta} = \dfrac{\cos^{2}\theta}{\cos^{2}\theta} = 1$
4. Ans. Option (C) is correct.
$\tan 60^{\circ} = \dfrac{PR}{PS} \Rightarrow \sqrt{3} = \dfrac{8}{PS} \Rightarrow PS = \dfrac{8}{\sqrt{3}} = \dfrac{8\sqrt{3}}{3}$ cm
5. Ans. Option (B) is correct.
$\tan 30^{\circ}\cot 60^{\circ} = \dfrac{1}{\sqrt{3}} \times \dfrac{1}{\sqrt{3}} = \dfrac{1}{3}$
Case Study — Read the following text and answer the following questions.
In the given figure, ABC, ADE and AQP are three right triangles.
(i) The value of $\sin A$ is the greatest for triangle PQA. Do you agree?
(ii) ABC is an isosceles right triangle, right-angled at B. What is the value of $2\sin A \times \cos A$?
OR
In $\triangle ADE$, $AE = 5$ cm and $AD = 3$ cm. Find the value of $2\sin A\cos A$.
(iii) Write one statement about trigonometric ratios in a right triangle.
In the given figure, ABC, ADE and AQP are three right triangles.
(i) The value of $\sin A$ is the greatest for triangle PQA. Do you agree?
(ii) ABC is an isosceles right triangle, right-angled at B. What is the value of $2\sin A \times \cos A$?
OR
In $\triangle ADE$, $AE = 5$ cm and $AD = 3$ cm. Find the value of $2\sin A\cos A$.
(iii) Write one statement about trigonometric ratios in a right triangle.
Answer
(i) No, all three triangles have the same value of $\sin A$. The value of $\sin A$ does not depend on the size of the triangle but is a ratio of specific side lengths — it remains constant for the same angle $A$ regardless of which triangle is considered.
(ii) [Main part — Isosceles right triangle ABC:]
Let $AB = BC = x$. By Pythagoras theorem: $AC = \sqrt{x^{2} + x^{2}} = x\sqrt{2}$
$$2\sin A \times \cos A = 2 \times \frac{x}{x\sqrt{2}} \times \frac{x}{x\sqrt{2}} = 2 \times \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} = \mathbf{1}$$ [OR — Triangle ADE:]
$AE = 5$ cm, $AD = 3$ cm. By Pythagoras: $DE = \sqrt{5^{2} – 3^{2}} = \sqrt{16} = 4$ cm
$$2\sin A\cos A = 2 \times \frac{4}{5} \times \frac{3}{5} = \frac{24}{25}$$ (iii) The values of $\sin A$ and $\cos A$ always lie between 0 and 1 (inclusive) for any acute angle $A$ in a right triangle.
(ii) [Main part — Isosceles right triangle ABC:]
Let $AB = BC = x$. By Pythagoras theorem: $AC = \sqrt{x^{2} + x^{2}} = x\sqrt{2}$
$$2\sin A \times \cos A = 2 \times \frac{x}{x\sqrt{2}} \times \frac{x}{x\sqrt{2}} = 2 \times \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} = \mathbf{1}$$ [OR — Triangle ADE:]
$AE = 5$ cm, $AD = 3$ cm. By Pythagoras: $DE = \sqrt{5^{2} – 3^{2}} = \sqrt{16} = 4$ cm
$$2\sin A\cos A = 2 \times \frac{4}{5} \times \frac{3}{5} = \frac{24}{25}$$ (iii) The values of $\sin A$ and $\cos A$ always lie between 0 and 1 (inclusive) for any acute angle $A$ in a right triangle.
Frequently Asked Questions
What does the Introduction to Trigonometry chapter cover in CBSE Class 10 Maths? ⌄
The Introduction to Trigonometry chapter in CBSE Class 10 covers the six trigonometric ratios (sine, cosine, tangent and their reciprocals), their values for standard angles (0°, 30°, 45°, 60°, 90°), and the three fundamental trigonometric identities. Students also learn to prove identities and simplify expressions — skills that are heavily tested across all question types in the board exam.
How many marks does the Introduction to Trigonometry chapter carry in the CBSE Class 10 board exam? ⌄
Introduction to Trigonometry falls under the Trigonometry unit in CBSE Class 10, which carries approximately 12 marks in the board paper. Questions appear as 1-mark MCQs, 2-mark short answers, 3-mark proofs, and 5-mark case studies — making it one of the most consistently high-scoring chapters when your child practises regularly.
What are the most important topics in Introduction to Trigonometry for board exams? ⌄
The most frequently tested topics are the three Pythagorean identities ($\sin^{2}\theta + \cos^{2}\theta = 1$, $\sec^{2}\theta – \tan^{2}\theta = 1$, and $\operatorname{cosec}^{2}\theta – \cot^{2}\theta = 1$), finding a ratio given another ratio, evaluating expressions using standard angle values, and proving trigonometric identities. Identity-based proof questions of 3 and 5 marks are especially common in previous year papers.
What common mistakes do students make in trigonometry questions? ⌄
A very common error is confusing $\tan^{2}\theta – \sec^{2}\theta = 1$ (which is wrong) with $\sec^{2}\theta – \tan^{2}\theta = 1$ (correct). Students also frequently mix up $\sin\theta$ and $\operatorname{cosec}\theta$, forget to simplify $\sin^{2}\theta + \cos^{2}\theta$ to 1 during proofs, and lose marks by skipping intermediate steps. Practising board paper proofs with fully shown steps helps your child avoid these errors under exam pressure.
How does Angle Belearn help students score well in Introduction to Trigonometry? ⌄
Angle Belearn’s CBSE specialists curate chapter-wise question banks drawn from real board papers, each paired with clear, step-by-step solutions that show exactly how marks are earned. Regular practice on these verified questions builds your child’s confidence with identity proofs, standard angle substitutions, and MCQ shortcuts — so they walk into the exam ready to score full marks across all question types.
