Question

Prove that:
(sinA+secA)^2+(cosA+cosecA)^2 = (1+cosecAsecA)^2​

Answer 1

Solution:

(sin A + sec A)² + (cos A + cosec A)² = (1 + cosec A sec A)²

We take LHS,

=> (sin A + sec A)² + (cos A + cosec A)²

=> sin²A + sec²A + 2 sin A sec A + cos²A + cosec²A + 2 cos A cosec A

We know that sec A = 1/cos A and cosec A = 1/sin A. So,

=> (sin²A + cos²A) + (sec²A + cosec²A) + $\sf{\bigg(\dfrac{2\sin\;A}{\cos\;A}+\dfrac{2\cos\;A}{\sin\;A}\bigg)}$

$\sf{\implies 1+\bigg(\dfrac{1}{\cos^{2}A}+\dfrac{1}{\sin^{2}A}\bigg)+\bigg(\dfrac{2\sin^{2}A+2\cos^{2}A}{\sin\;A\cos\;A}\bigg)}$

$\sf{\implies 1+\bigg(\dfrac{sin^{2}A+\cos^{2}A}{sin^{2}A\;cos^{2}A}\bigg)+\bigg[\dfrac{2(sin^{2}A+\cos^{2}A)}{sinA\;cosA}\bigg]}$

$\sf{\implies 1+\dfrac{1}{\sin^{2} A\; \cos^{2} A} + \dfrac{2}{\sin A\;\cos A}}$

$\sf{\implies \Bigg(1+\dfrac{1}{\sin A\;\cos A}\Bigg)^{2}}$

$\sf{\implies (1+cosec A \;\sec A)^{2}}$

LHS = RHS

Hence Proved!!!

Answer 2

$\huge\underline\green{\underline{Solution\:is\: attached\:here:-}}$