Question

prove that sin A-cos A+1/sin A +cos A -1=1/Sec A-Tan A

Answer 1

Answer:

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Answer 2

Concept

The formulas will use to solve the problem.

1. $\frac{\sin x}{\cos x}=\tan x$

2. $\frac{1}{cosx}= secx$

3. $Sec^2x-tan^2x=1$

Given information

Given  $\frac{sinA-cos A+1}{sin A +cos A -1} =\frac{1}{Sec A-tan A}$ we have to prove it.

Solution

Start with L.H.S

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

Dividing numerator  and denominator by cosA

$\frac{\tan A-1+\sec A}{\tan A+1-\sec A}=\frac{\tan A+\sec A-1}{\tan A-\sec A+1} \\&=\frac{\tan A+\sec A-\left[\sec A^{2} \tan A^{2}\right)}{\tan A-\sec A+1} \\&=\frac{\tan +\sec A-(\sec A-\tan A)(\sec A+\tan A 1)}{\tan A-\sec A+1} \\&=\frac{(\tan A+\sec A)[1-\sec A+\tan A]}{\tan A-\sec A+1} \\&=\frac{(\tan A+\sec A)(\sec A-\tan A)}{\sec A-\tan A} \\=\frac{\sec 2 A-\tan 2 A}{\sec A-\tan A}\\=\frac{1}{\sec A-\tan A}$

=RHS