Question

plzzzzzz help me out this question is so hard

Answer 1

Given:   $x+y=90^\circ$

To prove:   $\dfrac{\sin^2x}{\cos^2y}+\dfrac{\tan x\cdot \tan y+\tan x\cdot \cot y}{\sin x\cdot\sec y}=1+\text{cosec}^2 y$

Formula:   $\sin(90-\theta)=\cos\theta$

Proof:

Taking left side and to prove right side.

$\Rightarrow \dfrac{\sin^2x}{\cos^2y}+\dfrac{\tan x\cdot \tan y+\tan x\cdot \cot y}{\sin x\cdot\sec y}$

$\Rightarrow \dfrac{\sin^2(90-y)}{\cos^2y}+\dfrac{\tan x\cdot \tan y+\tan x\cdot \cot y}{\sin (90-y)\cdot\sec y}$

$\Rightarrow \dfrac{\cos^2y}{\cos^2y}+\dfrac{\tan x( \tan y+\cot y)}{\cos y\cdot\sec y}$

$\Rightarrow 1+\dfrac{\tan (90-y)( 1+\cot^2 y)}{\cot y\cdot \cos y\cdot\sec y}$

$\Rightarrow 1+\dfrac{\cot y\cdot \text{cosec}^2y}{\cot y}$

$\Rightarrow 1+ \text{cosec}^2y=RHS$

Hence Proved