Question

the answer of the attached photo with the steps​

Answer 1

ANSWER:-

Given:

A + B= 90°

To prove:

$\sqrt{ \frac{tan \: A \: tan \: B + tan \: A \: cot \: B}{sin \: A \: sec \: B} - \frac{ {sin}^{2}B }{ {cos}^{2} A} }$

Proof:

A + B = 90°

=) B= 90° - A.........(1)

Take L.H.S.

$= > \sqrt{ \frac{tan \: A \: tan \: B+ tan \: A \:cot \: B}{sin \: A \: sec \: B} - \frac{ {sin}^{2}B }{ {cos}^{2} A} } \\ \\ = > \sqrt{ \frac{tan \: A\: tan(90 -A) + tan \: A \: cot \: (90 - A)}{sin \: A \: sec \: (90 - A)} - \frac{ {sin}^{2} (90 - A)}{ {cos}^{2}A } } \\ \\ = > \sqrt{ \frac{tan \: A \: cot \: A\: + tan \: A \: . tan \: A}{sin \: A.cosec \: A} - \frac{ {cos}^{2} A }{ {cos}^{2} A } } \\ \\ = > \sqrt{ \frac{1 + {tan}^{2} A}{1} - 1 } \\ \\ = > \sqrt{ {tan}^{2}A} \\ \\ = > tan \: A \: \: \: \: R.H.S$

Hence,

Proved.

Hope it helps ☺️