Question

please solve this question... ​

Answer 1

EXPLANATION.

$\sf \implies A + B = 90. - - - - - (1).$

As we know that,

We can write equation as,

$\sf \implies B = 90 - A. - - - - - (1).$

$\sf \implies \sqrt{\dfrac{tan(A).tan(B) + tan(A).cot(B)}{sin(A).sec(B). } \ - \dfrac{sin^{2} B}{cos^{2} A}$

Put the value of B = 90 - A in equation, we get.

$\sf \implies \sqrt{\dfrac{tan(A).tan(90 - A) + tan(A).cot(90 - A)}{sin(A). sec(90 - A)} \ - \dfrac{sin^{2} (90 - A)}{cos^{2} A} }$

$\sf \implies \sqrt{\dfrac{tan(A).cot(A) + tan(A).tan(A)}{sin(A).cosec(A)} \ - \dfrac{cos^{2} A}{cos^{2}A } }$

$\sf \implies \sqrt{\dfrac{tan(A). \bigg(\dfrac{1}{tan(A)} \bigg) + tan^{2}A }{sin(A) . \bigg(\dfrac{1}{sin(A)}\bigg) } \ - 1}$

$\sf \implies \sqrt{\dfrac{1 + tan^{2}A }{1} \ - 1}$

$\sf \implies \sqrt{1 + tan^{2} A - 1} \ = \sqrt{tan^{2} A} \ = tan(A).$

Hence Proved.