Question

please answer this question without using latex ?

Class - 10th
Chapter - Trignometry ​

Answer 1

Given:-

$\frac{sin \: x - cos \: x + 1}{sin \: x + cos \: x - 1} = \frac{1}{sec \: x - tan \: x}$

Solution:-

LHS=

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

Dividing by cos, we get,

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

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

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

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

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

$= sec \: x + tan \: x \times \frac{sec \: x - tan \: x}{sec \: x - tan \: x}$

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

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

= RHS

Hence, proved.

Answer 2

Step-by-step explanation:

Please. refer attachment

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