SSC CHSL 2024 PYQ — Question 71

The correct answer is option 3 i.e. 2 $cos\theta$.

$\left(\frac{sin\theta}{1+cos\theta}+\frac{1+cos\theta}{sin\theta}\right)\left(\frac{1}{tan\theta + cot\theta}\right)$

$=\left(\frac{sin^2\theta + (1+cos\theta)^2}{sin\theta(1+cos\theta)} \right)\left(\frac{1}{\frac{sin\theta}{cos\theta} + \frac{cos\theta}{sin\theta}}\right)$

$=\left(\frac{sin^2\theta + 1 + 2cos\theta + cos^2\theta}{sin\theta(1+cos\theta)}\right)\left(\frac{1}{\frac{sin^2\theta + cos^2\theta}{sin\theta cos\theta}}\right)$

$=\left(\frac{1 + 1 + 2cos\theta}{sin\theta(1+cos\theta)}\right)\left(\frac{1}{\frac{1}{sin\theta cos\theta}}\right)$

$=\left(\frac{2(1 + cos\theta}{sin\theta(1+cos\theta)}\right)\left(sin\theta cos\theta\right)$

= (2/sinθ)(sinθ cosθ)

= 2cosθ

Therefore, the value of the expression is 2cosθ.