Question

Prove that :
sinA - cos A +1/ sinA + cos A -1=
1/secA - tanA​

Answer 1

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Hope it will help u friend...

In 4th step it is written sec^2A-tan^2A instead of 1...

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Answer 2

Answer with Step-by-step explanation:

Given:

LHS

$\frac{sinA-cosA+1}{sinA+cosA-1}$

Divide numerator and denominator with cos A

$\frac{\frac{sinA}{cosA}-\frac{cosA}{cosA}+\frac{1}{cosA}}{\frac{sinA}{cosA}+\frac{cosA}{cosA}-\frac{1}{cosA}}$

$=\frac{tanA-1+secA}{tanA+1-secA}$

Using the formula

$tanA=\frac{sinA}{cosA}$

$secA=\frac{1}{cosA}$

$\frac{tanA+secA-1}{tanA-secA+(sec^2A-tan^2A)}$

Using the identity

$sec^2A-tan^2A=1$

$\frac{tanA+secA-1}{-1(-tanA+secA)+(secA-tanA)(secA+tanA)}$

$\frac{tanA+secA-1}{secA-tanA(secA+tanA-1)}$

$\frac{1}{secA-tanA}$

LHS=RHS

Hence, proved.

#Learn more:

https://brainly.in/question/8593084:Answered by Piyu