Question

Prove that :-

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

Answer 1

To prove:

$\frac{\sin A-\cos A+1}{\sin A+\cos A-1}=$L.H.S$=\frac{\cos A}{1-\sin A}=$R.H.S

L.H.S$=\frac{\sin A+(1-\cos A)}{\sin A-(1-\cos A)}\to 1$

We know,

$\cos 2\theta =2{\cos }^2\theta -1$

$=1-2{\sin }^2\theta$

$\Rightarrow 1-\cos 2\theta =2{\sin }^2\theta$

$\Rightarrow 1-\cos A=2{\sin }^2\frac{A}{2}$

Putting this value of $(1-\cos A)$ in $1$

$\Rightarrow$L.H.S$=\frac{\sin A+2{\sin }^2\frac{A}{2}}{\sin A-2{\sin }^2\frac{A}{2}}\to 2$

We also know $\sin 2\theta =2\sin \theta \cos \theta$

$\Rightarrow \sin A=2\sin \frac{A}{2}\cos \frac{A}{2}$

Putting this value of $\sin A$ in $2$

L.H.S$=\frac{2\sin \frac{A}{2}\cos \frac{A}{2}+2{\sin }^2\frac{A}{2}}{2\sin \frac{A}{2}\cos \frac{A}{2}-2{\sin }^2\frac{A}{2}}$

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

$=\frac{\cos \frac{A}{2}+\sin \frac{A}{2}}{\cos \frac{A}{2}-\sin \frac{A}{2}}$

Rationalizing by multiplying and dividing the above term with denominator i.e., $\cos \frac{A}{2}-\sin \frac{A}{2}$

$=\left(\frac{\cos \frac{A}{2}+\sin \frac{A}{2}}{\cos \frac{A}{2}-\sin \frac{A}{2}}\right)\times \left(\frac{\cos \frac{A}{2}-\sin \frac{A}{2}}{\cos \frac{A}{2}-\sin \frac{A}{2}}\right)$

$=\frac{{\cos }^2\frac{A}{2}-{\sin }^2\frac{A}{2}}{{(\cos \frac{A}{2}-\sin \frac{A}{2})}^2}\to 3$

Now, ${\cos }^2\theta -{\sin }^2\theta =\cos 2\theta$

$\Rightarrow {\cos }^2\frac{A}{2}-{\sin }^2\frac{A}{2}=\cos A$

Putting in $3$

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

$=\frac{\cos A}{1-2\sin \frac{A}{2}\cos \frac{A}{2}}$

$=\frac{\cos A}{1-\sin A}=$R.H.S

Hence proved

Formulae used in question

$\cdot \cos 2\theta =2{\cos }^2\theta -1=1-2{\sin }^2\theta$

$\cdot \cos 2\theta ={\cos }^2\theta -{\sin }^2\theta$

$\cdot \sin 2\theta =2\sin \theta \cos \theta$

$\cdot {\sin }^2\theta +{\cos }^2\theta =1$