Question

solve.............................

Answer 1

Given : tan 15 + cot 75

= > tan 15 + tan(90 - 15)

We know that tan(90 - A) = cotA

= > tan 15 + cot 15

We know that cot A = (1/tanA)

$=> tan 15 + \frac{1}{tan 15}$

$=> \frac{tan^2 15 + 1}{tan 15}$

We know that sec^2x - tan^2x = 1 (or) tan^2x + 1 = sec^2x

$=> \frac{sec^2 15}{tan 15}$

$=> \frac{sec^2 15}{\sqrt{tan^2 15} }$

$=> \frac{sec^2 15}{\sqrt{sec^2 15 - 1} }$

Hope this helps!

Answer 2

HI HERE IS THE ANSWER