SSC CHSL 2021 PYQ — Question 64

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

Given:

3 cotA = 4 tanA

Formula Used:

tanA = $P \over B$

By pythagoras theorem, H² = P² + B²

Explanation:

According to the given question,

3 cotA = 4 tanA

⇒ tan²A = $3 \over 4$

⇒ tanA = $\sqrt{3} \over 2$

Now, P = $\sqrt{3}$ ; B = 2; H = $\sqrt{7}$

Now, secA = $\sqrt{7}\over 2$

Hence the required answer is $\sqrt{7}\over 2$.