Question

Prove that tan A+ sec A-1 / tan A - sec A+1 = 1+sin A/cos A

Answer 1

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Answer 2

To Prove:

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

Solution:    

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

$\left[\because 1+\tan ^{2} A=\sec ^{2} A \Rightarrow \sec ^{2} A-\tan ^{2} A=1\right]$

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

Now, take$(\tan A+\sec A)$ as a common term, we get

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

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

$=\tan A+\sec A$

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

$=\frac{1+\sin A}{\cos A}$    

∴ Hence Proved