Question

guys,,, please solve this question....

Answer 1

$\mathfrak{\huge{\green{\underline{\red{Hola\: !}}}}}$

$\mathbb{\underline{\blue{SOLUTION:}}}$

taking theta as x

$\mathcal{\underline{\purple{L.H.S = }}}$

$\frac{sin x - cos x + 1}{sin x + cos x - 1}$

Divide Numerator and Denominator by cos x

=> $\frac{\frac{sin x}{cos x} - \frac{cos x}{cos x} + \frac{1}{cos x}}{\frac{sin x}{cos x} - \frac{cos x}{cos x} - \frac{1}{cos x}}$

=> $\frac{tan x - 1 + sec x}{tan x + 1 - sec x}$

=> $\frac{tan x +sec x - 1}{tan x - sec x + 1}$

=> $\frac{tanx + sec x - (sec^{2} x - tan^{2} x)}{tan x - sec x + 1)}$

=> $\frac{tan x + sec x - (sec^{2} x - tan^{2} x)}{tan x - sec x + 1}$

=> $\frac{tan x + sec x - ((sec x + tan x)(sec x - tan x))}{tan x - sec x + 1}$

=> $\frac{tan x + sec x - (tan x + sec x)(sec x - tan x)}{tan x - sec x + 1}$

=> $\frac{(tan x + sec x)[1 -(sec x - tan x)]}{1 - sec x + tan x}$

=> $\frac{(tan x + sec x)[1 - sec x + tan x)]}{1 - sec x + tan x }$

=> sec x + tan x

Rationalise

=> $\frac{(sec x + tan x)(sec x - tan x)}{sec x - tan x}$

=> $\frac{sec^{2} x - tan^{2}x}{sec x - tan x}$

=> $\frac{1}{sec x - tan x}$

$\mathcal{\underline{\purple{= R.H.S}}}$

$\underline{\underline{Hence\: Proved}}$

$\mathbb{\underline{\purple{Identities\: Used : }}}$

•$(a - b)^{2} = (a+b)(a-b)$
[polynomial Identity]

•$sec^{2} x - tan^{2} x = 1$
[Trigonometric Identity]

$\mathfrak{\huge{\pink{Cheers}}}$

$\mathcal{\huge{\orange{Hope\: it\: Helps}}}$

Answer 2

Hi friend hope this attachment will help you.....