The value of $\frac{\sin^2 48^\circ + \sin^2 42^\circ}{\tan^2 40^\circ - \csc^2 50^\circ}$ is:

The correct answer is Option A: 1. Use complementary angle identities: replace $\sin^2 42^\circ$ with $\cos^2 48^\circ$ to get $1$ in the numerator, and recognize the denominator as $-(1 + \cot^2 50^\circ - \tan^2 40^\circ) = -1$ since $\tan 40^\circ = \cot 50^\circ$. Step-by-Step Solution & Concepts: Shortcut Trick: Use complementary angle identities: replace $\sin^2 42^\circ$ with $\cos^2 48^\circ$ to get $1$ in the numerator, and recognize the denominator as $-(1 + \cot^2 50^\circ - \tan^2 40^\circ) = -1$ since $\tan 40^\circ = \cot 50^\circ$. Formula Used: $$\sin(90^\circ - \theta) = \cos \theta$$, $$\sin^2 \theta + \cos^2 \theta = 1$$, $$\tan(90^\circ - \theta) = \cot \theta$$, $$1 + \cot^2 \theta = \operatorname{cosec}^2 \theta$$ Given: Expression to evaluate is \frac{\sin^2 48^\circ + \sin^2 42^\circ}{\tan^2 40^\circ - \operatorname{cosec}^2 50^\circ} Steps: Simplify Numerator: Using the identity $\sin^2 \theta + \cos^2 \theta = 1$, we convert $\sin^2 42^\circ$ to $\cos^2 48^\circ$ since $42^\circ = 90^\circ - 48^\circ$.. Simplify Denominator: Given the denominator $\tan^2 40^\circ - \operatorname{cosec}^2 50^\circ$, we use $\tan 40^\circ = \cot 50^\circ$ and the identity $1 + \cot^2 \theta = \operatorname{cosec}^2 \theta$.. Calculate Final Result: Divide the numerator by the denominator. Answer: C is correct because the evaluated expression yields the result of -1. Exam Tip: These types of problems frequently appear in competitive exams to test your ability to quickly identify and apply complementary angle properties ($90^\circ - \theta$) and basic trigonometric identities.

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The value of $\frac{\sin^2 48°+ \sin^2 42°{\tan^2 40°- \csc^2 50°$ is:

Correct Answer

📐 MA • Math Trigonometric IdentityMathematicsPrevious Year Paper 2024

Option A

Quick Summary:

Use complementary angle identities: replace $\sin^2 42°$ with $\cos^2 48°$ to get $1$ in the numerator, and recognize the denominator as $-(1 + \cot^2 50°- \tan^2 40°) = -1$ since $\tan 40°= \cot 50°$ .

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📐MAMath SolutionTrigonometric Identity

📋 Given

Expression to evaluate is $\frac$ { $\sin$ ² 48°+ $\sin$ ² 42°{ $\tan$ ² 40°- \operatorname{cosec}^2 50°

🔢 Formula Used

sin⁡(90°−θ)=cos⁡θ\sin(90°- \theta) = \cos \thetasin(90°−θ)=cosθ, sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1sin2θ+cos2θ=1, tan⁡(90°−θ)=cot⁡θ\tan(90°- \theta) = \cot \thetatan(90°−θ)=cotθ, 1+cot⁡2θ=cosec⁡2θ1 + \cot^2 \theta = \operatorname{cosec}^2 \theta1+cot2θ=cosec2θ

⚡ Exam Hall Shortcut / Speed Trick

⚠️ Common Student Trap / Pitfall

Confusing signs in the denominator. The denominator is $\tan^2 40°- \operatorname{cosec}^2 50°$ . Using $\tan 40°= \cot 50°$ , this becomes $\cot^2 50°- (1 + \cot^2 50°) = -1$ .

📊 Diagram / Illustration

🔢 Step-by-Step Solution

Simplify Numerator

Using the identity $\sin^2 \theta + \cos^2 \theta = 1$ , we convert $\sin^2 42°$ to $\cos^2 48°$ since $42°= 90°- 48°$ .

sin⁡248°+sin⁡242°=sin⁡248°+cos⁡248°=1\sin^2 48°+ \sin^2 42°= \sin^2 48°+ \cos^2 48°= 1sin248°+sin242°=sin248°+cos248°=1

Simplify Denominator

Given the denominator $\tan^2 40°- \operatorname{cosec}^2 50°$ , we use $\tan 40°= \cot 50°$ and the identity $1 + \cot^2 \theta = \operatorname{cosec}^2 \theta$ .

cot⁡250°−cosec⁡250°=−1\cot^2 50°- \operatorname{cosec}^2 50°= -1cot250°−cosec250°=−1

Calculate Final Result

Divide the numerator by the denominator.

1−1=−1\frac{1}{-1} = -1−11​=−1

✅

C is correct because the evaluated expression yields the result of -1.

Core Concepts Used

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Trigonometric Ratios Complementary Angles Pythagorean Identities

💡 EXAM TIP

These types of problems frequently appear in competitive exams to test your ability to quickly identify and apply complementary angle properties ( $90°- \theta$ ) and basic trigonometric identities.

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