Prove the given identity, where the angles involved are acute ang… - Vidyadip

Question

Prove the given identity, where the angles involved are acute angles for which the expressions are defined.$\frac{\cos A-\sin A+1}{\cos A+\sin A-1}$ = cosec A + cot A, using the identity $cosec^2 A = 1 + cot^2 A$

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Answer

Taking L.H.S
$\frac{\cos A-\sin A+1}{\cos A+\sin A-1}$
Dividing Numerator and Denominator by sin A
$=\frac{\frac{\cos A-\sin A+1}{\sin A}}{\frac{\cos A+\sin A-1}{\sin A}}$
$=\frac{\frac{\cos A}{\sin A}-\frac{\sin A}{\sin A}+\frac{1}{\sin A}}{\frac{\cos A}{\sin A}+\frac{\sin A}{\sin A}-\frac{1}{\sin A}}$
Using the formula $\cot \theta = \frac{{\cos \theta }}{{\sin \theta }}$$=\frac{\cot A-1+\ cosec A}{\cot A+1-\ cosec A}$
Using the identity $cosec^2A = 1 + cot^2A$
$=\frac{\cot A-\left(\ cosec ^{2} A-\cot ^{2} A\right)+\ cosec A}{\cot A+1-\ cosec A}$
$=\frac{(\cot A+\ cosec A)-\left(\ cosec ^{2} A-\cot ^{2} A\right)}{\cot A+1-\ cosec A}$
$=\frac{(\cot A+\ cosec A)(1-\ cosec A+\cot A)}{\cot A+1-\ cosec A}$
$= cot A + cosec A$
$= R.H.S$

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