Question

Prove that: $[\frac{\sin A}{1-cosA}-1-\frac{\cos A}{\sin A}][\frac{\cos A}{1}+\frac{\sin A}{\cos A}]=4\csc A.$?

Answer 1

$LHS=[\frac{\sin A}{1-\cos A}-\frac{1-\cos A}{\sin A}][\frac{\cos A}{1+\sin A}+\frac{1+\sin A}{\cos A}]$

Now $1st$ part $=\frac{\sin A}{1-\cos A}-\frac{1-\cos A}{\sin A}$

$=\frac{\sin A(1+\cos A)}{(1-\cos A)(1+\cos A)}-\frac{1-\cos A}{\sin A}$

$=\frac{\sin A(1+\cos A)}{1-{\cos }^{2}A}-(\frac{1}{\sin A}-\frac{\cos A}{\sin A})$

$=\frac{\sin A(1+\cos A)}{{\sin }^{2}A}-\csc A+\cot A$

$=\frac{1+\cos A}{\sin A}-\csc A+\cot A$

$=\frac{1}{\sin A}+\frac{\cos A}{\sin A}-\csc A+\cot A$

$=\csc A+\cot A-\csc A+\cot A$

$=2\cot A$

$2nd$ part

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

$=\frac{\cos A(1-\sin A)}{(1+\sin A)(1-\sin A)}+(\frac{1}{\cos A}+\frac{\sin A}{\cos A})$

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

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

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

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

$=\sec A-\tan A+\sec A+\tan A$

$=2\sec A$

Whole $LHS$

$=2\cot A\times 2\sec A$

$=\frac{4\cos A}{\sin A}\times \frac{1}{\cos A}$

$=\frac{4}{\sin A}=4\csc A=RHS$

Proved