SSC CGL 2021 PYQ — Question 75

The correct answer is option 2 i.e. $-\frac{18}{7}$

If   tan $R = \frac{1}{3}$ = P/B

(Hypotenuse)² = 3² + 1²

H = $\sqrt{10}$.

sinP = 3/$\sqrt{10}$ and sinR = 1/$\sqrt{10}$

cosR = 3/$\sqrt{10}$ 

cosecP =$\sqrt{10}$/3 and cosecR =$\sqrt{10}$/1

secP = $\sqrt{10}$

⇒ $\frac{secP(cosR+sinP)}{cosecR(sinR-cosecP)}$ = $\frac{\sqrt{10}(\frac{3}{\sqrt{10}}+\frac{3}{\sqrt{10}})}{\sqrt{10}(\frac{1}{\sqrt{10}}- \frac{\sqrt{10}}{3})}$

⇒ $\frac{6 \times 3\sqrt{10}}{\sqrt{10} \times -7}$= -18/7.