SSC CHSL 2021 PYQ — Question 70

The correct answer is option 2 i.e. $5\frac{2}{3}$.

3cos²$\theta$ - 4sin$\theta$ + 1 = 0

⇒ 3( 1 -  sin²$\theta$ ) - 4sin$\theta$ + 1 = 0

⇒ 3 -  3sin²$\theta$ - 4sin$\theta$ + 1 = 0

⇒ 3sin²$\theta$  +  4sin$\theta$ - 4 = 0

⇒ 3sin²$\theta$ + 6sin$\theta$ - 2sin$\theta$ - 4 = 0

⇒ 3sin$\theta$ (sin$\theta$  + 2) - 2(sin$\theta$ + 2) = 0

⇒ (3 sin$\theta$  - 2)( sin$\theta$ + 2) = 0

⇒ sin$\theta$ = $2 \over 3$;  sin$\theta$ = -2 (Not possible)

Now, P = 2 and H = 3

So, B = $\sqrt{5}$

Then,  3cos²$\theta$ + 5tan² $\theta$ = $3 \times {5 \over 9} + 5 \times {4 \over 5}$ = ${5 \over 3} + 4$ = $5\frac{2}{3}$

Hence, the required answer is $5\frac{2}{3}$.