SSC CGL 2023 PYQ — Question 70

The correct answer is Option 3 i.e. 0.

2 cotθ = 3

⇒ cotθ = 3/2 = B/P

⇒ H² = B² + P²

⇒ H² = 3² + 2²

⇒ H² = 9 + 4

⇒ H² = 13

⇒ H = √13

Now, sinθ = P/H = 2/√13

⇒ tanθ = P/B = 2/3

⇒ cosθ = B/H = 3/√13

Now,

$\frac{\sqrt{13}sin\theta-3tan\theta }{3tan\theta+ \sqrt{13}cos\theta}$

⇒ $\frac{\sqrt{13}\times {2\over \sqrt13 }-3\times {2\over3}}{3\times {2\over3}+ \sqrt{13}\times{3\over \sqrt 13}}$

⇒ $\frac{{2 - 2}}{3\times {2\over3}+ \sqrt{13}\times{3\over \sqrt 13}}$ = 0