((tan³ θ)/(sec² θ)+(cot³ θ)/(cosec² θ)+2sin \ θ \ cos \ θ) ÷ (1 + cosec² θ + tan² θ), 0º < θ < 90º, is equal to:

((tan³ θ)/(sec² θ)+(cot³ θ)/(cosec² θ)+2sin \ θ \ cos \ θ)…

$(\frac{tan^3 \theta}{sec^2 \theta}+\frac{cot^3 \theta}{cosec^2 \theta}+2sin \ \theta \ cos \ \theta)$ ÷ (1 + cosec² θ + tan² θ), 0º < θ < 90º, is equal to:

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Correct answer: C

Explanation

The correct answer is option 3 i.e. sinθ cosθ

= $(\frac{tan^3 \theta}{sec^2 \theta}+\frac{cot^3 \theta}{cosec^2 \theta}+2sin \ \theta \ cos \ \theta)$ ÷ (1 + cosec² θ + tan² θ)

= $\frac{\frac{sin^3 \theta}{cos^3 \theta}}{\frac{1}{cos^2 \theta}}$ + $\frac{\frac{cos^3 \theta}{sin^3 \theta}}{\frac{1}{sin^2 \theta}}$ + 2sinθcosθ ÷ (cosec² θ + sec² θ)

= $\frac{ (sin^2 \theta)^2 + (cos^2 \theta)^2+ 2sin^2 \theta cos^2 \theta}{sin \theta cos \theta}$ ÷ (sin² θ + cos² θ)/(sin²θ cos²θ)

= sinθ cosθ.

SSC CGL 2022 PYQ — Question 20