20. Trigonometrical Identities Previous Year Questions

From sin²A + sin A = 1 ⇒ sin A = 1 − sin²A = cos²A. Squaring gives sin²A = cos⁴A ⇒ 1 − cos²A = cos⁴A ⇒ cos⁴A + cos²A = 1. Both statements true and Reason is correct justification. [1 Mark]

Denominator: sin³A + cos³A = (sin A + cos A)(sin²A − sin A cos A + cos²A) = (sin A + cos A)(1 − sin A cos A). [1.5 Marks]
Numerator factor: cot A + tan A − 1 = (cos A/sin A + sin A/cos A) − 1 = (1 / sin A cos A) − 1 = (1 − sin A cos A) / (sin A cos A). [1.5 Marks]
Cancelling (sin A + cos A)(1 − sin A cos A) gives 1 / (sin A cos A) = sec A · csc A = RHS. Hence Proved. [1 Mark]

Correct Option: (a) only 1
sin²θ + cos²θ = 1 is valid. The second is invalid because cosec²θ − cot²θ = 1. [1 Mark]

sin³θ + cos³θ = (sinθ + cosθ)(sin²θ − sinθcosθ + cos²θ) = (sinθ + cosθ)(1 − sinθcosθ). [2 Marks]
LHS = [ (sinθ+cosθ)(1−sinθcosθ) / (sinθ+cosθ) ] + sinθcosθ = (1 − sinθcosθ) + sinθcosθ = 1 = RHS. [2 Marks]

Using identity $(a + b)(a - b) = a^2 - b^2$:
$(1 + \sin A)(1 - \sin A) = 1 - \sin^2 A$.
Using fundamental Pythagorean identity $\sin^2 A + \cos^2 A = 1 \implies 1 - \sin^2 A = \cos^2 A$.
Correct Option: (d) [1 Mark]

Step 1 State standard Pythagorean identities:
$\sin^2 \theta + \cos^2 \theta = 1 \implies \sin^2 \theta - 1 = -\cos^2 \theta$.1 Mark
$1 + \tan^2 \theta = \sec^2 \theta$.1 Mark
Step 2 Substitute into LHS:
$\text{LHS} = (-\cos^2 \theta)(\sec^2 \theta) + 1$.1 Mark
Step 3 Since $\sec \theta = \frac{1}{\cos \theta}$, $\cos^2 \theta \cdot \sec^2 \theta = 1$:
$\text{LHS} = -(1) + 1 = 0 = \text{RHS}$.
Hence Proved!1 Mark

Step 1 Express $\cot A$ and $\tan A$ in terms of $\sin A$ and $\cos A$:
$$\text{LHS} = \frac{\sin A}{1 + \frac{\cos A}{\sin A}} - \frac{\cos A}{1 + \frac{\sin A}{\cos A}} = \frac{\sin A}{\frac{\sin A + \cos A}{\sin A}} - \frac{\cos A}{\frac{\cos A + \sin A}{\cos A}}$$ [1 Mark]
Step 2 Simplify the compound fractions:
$$= \frac{\sin^2 A}{\sin A + \cos A} - \frac{\cos^2 A}{\sin A + \cos A} = \frac{\sin^2 A - \cos^2 A}{\sin A + \cos A}$$ [1 Mark]
Step 3 Factor the difference of squares in the numerator:
$$= \frac{(\sin A - \cos A)(\sin A + \cos A)}{\sin A + \cos A} = \sin A - \cos A = \text{RHS}$$
Hence Proved. [1 Mark]

Step 1 Express $\cot \theta$ as $\frac{1}{\tan \theta}$:
$$\text{LHS} = \left(\frac{1 - \tan \theta}{1 - \frac{1}{\tan \theta}}\right)^2$$ [1.5 Marks]
Step 2 Simplify the denominator fraction:
$$1 - \frac{1}{\tan \theta} = \frac{\tan \theta - 1}{\tan \theta} = -\frac{1 - \tan \theta}{\tan \theta}$$ [1.5 Marks]
Step 3 Substitute back into LHS:
$$\text{LHS} = \left(\frac{1 - \tan \theta}{-\frac{1 - \tan \theta}{\tan \theta}}\right)^2 = \left(-\tan \theta\right)^2 = \mathbf{\tan^2 \theta} = \text{RHS}$$
Hence Proved. [1 Mark]

Step 1 Express all terms in terms of $\sin\theta$ and $\cos\theta$:
$\text{LHS} = \left(\frac{1}{\sin\theta} - \sin\theta\right) \left(\frac{1}{\cos\theta} - \cos\theta\right) \left(\frac{\sin\theta}{\cos\theta} + \frac{\cos\theta}{\sin\theta}\right)$. [1 Mark]
Step 2 Simplify each bracket:
$\text{Bracket 1} = \frac{1 - \sin^2\theta}{\sin\theta} = \frac{\cos^2\theta}{\sin\theta}$.
$\text{Bracket 2} = \frac{1 - \cos^2\theta}{\cos\theta} = \frac{\sin^2\theta}{\cos\theta}$.
$\text{Bracket 3} = \frac{\sin^2\theta + \cos^2\theta}{\sin\theta\cos\theta} = \frac{1}{\sin\theta\cos\theta}$. [1 Mark]
Step 3 Multiply the expressions:
$\text{LHS} = \left(\frac{\cos^2\theta}{\sin\theta}\right) \times \left(\frac{\sin^2\theta}{\cos\theta}\right) \times \left(\frac{1}{\sin\theta\cos\theta}\right) = \frac{\sin^2\theta\cos^2\theta}{\sin^2\theta\cos^2\theta} = \mathbf{1} = \text{RHS}$.
Hence Proved. [1 Mark]

Solution $\sec\theta (1 + \sin\theta) = \sec\theta + \sec\theta\sin\theta = \sec\theta + \tan\theta$.
Then $(\sec\theta + \tan\theta)(\sec\theta - \tan\theta) = \sec^2\theta - \tan^2\theta = 1$.
Hence Assertion (A) is true, Reason (R) is true, and (R) is the exact justification for (A).
Correct Answer: (c) Both (A) and (R) are true and (R) is the correct explanation of (A). [1 Mark]

Step 1 Combine the fractions by taking the common denominator:
$\text{LHS} = \frac{\left(\sqrt{\sec\theta + 1}\right)^2 + \left(\sqrt{\sec\theta - 1}\right)^2}{\sqrt{(\sec\theta - 1)(\sec\theta + 1)}} = \frac{(\sec\theta + 1) + (\sec\theta - 1)}{\sqrt{\sec^2\theta - 1}}$. [1.5 Marks]
Step 2 Simplify numerator and denominator:
$\text{Numerator} = 2\sec\theta$.
Using identity $\sec^2\theta - 1 = \tan^2\theta$, $\sqrt{\sec^2\theta - 1} = \tan\theta$.
$\text{LHS} = \frac{2\sec\theta}{\tan\theta}$. [1.5 Marks]
Step 3 Express in terms of sine and cosine:
$\frac{2\sec\theta}{\tan\theta} = 2 \times \frac{1/\cos\theta}{\sin\theta/\cos\theta} = \frac{2}{\sin\theta} = \mathbf{2\text{ cosec}\,\theta} = \text{RHS}$.
Hence Proved. [1 Mark]

Method 1 Using reciprocal relations:
$$\text{LHS} = \sqrt{\sec^2 \theta + \csc^2 \theta} = \sqrt{\frac{1}{\cos^2 \theta} + \frac{1}{\sin^2 \theta}}$$ [1 Mark]
$$\text{LHS} = \sqrt{\frac{\sin^2 \theta + \cos^2 \theta}{\sin^2 \theta \cos^2 \theta}} = \sqrt{\frac{1}{\sin^2 \theta \cos^2 \theta}} = \frac{1}{\sin \theta \cos \theta}$$ [1 Mark]
Express 1 in numerator as $\sin^2 \theta + \cos^2 \theta$:
$$\frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} = \frac{\sin^2 \theta}{\sin \theta \cos \theta} + \frac{\cos^2 \theta}{\sin \theta \cos \theta} = \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta} = \mathbf{\tan \theta + \cot \theta} = \text{RHS}$$ [1 Mark]
Hence Proved.

Step 1 Convert all terms into $\sin \theta$ and $\cos \theta$:
$$\text{LHS} = \left(1 + \frac{\cos \theta}{\sin \theta} - \frac{1}{\sin \theta}\right) \left(1 + \frac{\sin \theta}{\cos \theta} + \frac{1}{\cos \theta}\right)$$ [1 Mark]
Step 2 Take common denominators in each bracket:
$$\text{LHS} = \left(\frac{\sin \theta + \cos \theta - 1}{\sin \theta}\right) \left(\frac{\cos \theta + \sin \theta + 1}{\cos \theta}\right)$$ [1 Mark]
Step 3 Group $(\sin \theta + \cos \theta)$ as $a$ and $1$ as $b$ using $(a - b)(a + b) = a^2 - b^2$:
$$\text{Numerator} = (\sin \theta + \cos \theta)^2 - 1^2 = \sin^2 \theta + \cos^2 \theta + 2\sin \theta \cos \theta - 1$$ [1 Mark]
Since $\sin^2 \theta + \cos^2 \theta = 1$:
$$\text{Numerator} = 1 + 2\sin \theta \cos \theta - 1 = 2\sin \theta \cos \theta$$ $$\text{LHS} = \frac{2\sin \theta \cos \theta}{\sin \theta \cos \theta} = \mathbf{2} = \text{RHS}$$ [1 Mark]
Hence Proved.

Step 1 Complementary angles for $\sin$:
$$\sin 62^\circ = \sin(90^\circ - 28^\circ) = \cos 28^\circ$$ $$\sin^2 28^\circ + \sin^2 62^\circ = \sin^2 28^\circ + \cos^2 28^\circ = 1$$ [1 Mark]
Step 2 Complementary angles for $\cot$:
$$\cot 52^\circ = \cot(90^\circ - 38^\circ) = \tan 38^\circ$$ $$\tan^2 38^\circ - \cot^2 52^\circ = \tan^2 38^\circ - \tan^2 38^\circ = 0$$ [1 Mark]
Step 3 Standard angle $\sec 30^\circ = \frac{2}{\sqrt{3}}$:
$$\frac{1}{4} \sec^2 30^\circ = \frac{1}{4} \times \left(\frac{2}{\sqrt{3}}\right)^2 = \frac{1}{4} \times \frac{4}{3} = \frac{1}{3}$$ $$\text{Total Value} = 1 + 0 + \frac{1}{3} = \mathbf{\frac{4}{3}} = \mathbf{1\frac{1}{3}}$$ [1 Mark]
Answer: $\frac{4}{3}$ (or $1\frac{1}{3}$).

Step 1 Factor out $\sin \theta$ in numerator and $\cos \theta$ in denominator:
$$\text{LHS} = \frac{\sin \theta(1 - 2\sin^2 \theta)}{\cos \theta(2\cos^2 \theta - 1)}$$ [1 Mark]
Step 2 Express $1$ as $\sin^2 \theta + \cos^2 \theta$:
$$1 - 2\sin^2 \theta = \sin^2 \theta + \cos^2 \theta - 2\sin^2 \theta = \cos^2 \theta - \sin^2 \theta$$ $$2\cos^2 \theta - 1 = 2\cos^2 \theta - (\sin^2 \theta + \cos^2 \theta) = \cos^2 \theta - \sin^2 \theta$$ [1 Mark]
Step 3 Cancel common terms:
$$\text{LHS} = \frac{\sin \theta}{\cos \theta} \times \frac{\cos^2 \theta - \sin^2 \theta}{\cos^2 \theta - \sin^2 \theta} = \mathbf{\tan \theta} = \text{RHS}$$ [1 Mark]
Hence Proved.