Question

$\frac{\sqrt{1+\sin 7x}-\sqrt{1-\sin 7x}}{\sqrt{1-\sin 7x}+\sqrt{1-\sin 7x}}=\tan \frac{7x}{2}$

Answer 1

We have,

$\frac{\sqrt{1+\sin 7x}-\sqrt{1-\sin 7x}}{\sqrt{1+\sin 7x}+\sqrt{1-\sin 7x}}=\tan \frac{7x}{2}$

On reciprocal and we get,

$\frac{\sqrt{1+\sin 7x}+\sqrt{1-\sin 7x}}{\sqrt{1+\sin 7x}-\sqrt{1-\sin 7x}}=\cot \frac{7x}{2}$

Using componendo and dividendo rule

$\frac{\sqrt{1+\sin 7x}+\sqrt{1-\sin 7x}+\sqrt{1+\sin 7x}-\sqrt{1-\sin 7x}}{\sqrt{1+\sin 7x}+\sqrt{1-\sin 7x}-\sqrt{1+\sin 7x}+\sqrt{1-\sin 7x}}=\frac{\cot \frac{7x}{2}+1}{\cot \frac{7x}{2}-1}$

$\Rightarrow \frac{2\sqrt{1+\sin 7x}}{2\sqrt{1-\sin 7x}}=\frac{\frac{\cos \frac{7x}{2}}{sin\frac{7x}{2}}+1}{\frac{\cos \frac{7x}{2}}{sin\frac{7x}{2}}-1}$

$\Rightarrow \frac{\sqrt{1+\sin 7x}}{\sqrt{1-\sin 7x}}=\frac{\cos \frac{7x}{2}+sin\frac{7x}{2}}{\cos \frac{7x}{2}-sin\frac{7x}{2}}$

$\Rightarrow \frac{\sqrt{{\sin }^2\frac{7x}{2}+{\cos }^2\frac{7x}{2}+2\sin \frac{7x}{2}\cos \frac{7x}{2}}}{\sqrt{{\sin }^2\frac{7x}{2}+{\cos }^2\frac{7x}{2}-2\sin \frac{7x}{2}\cos \frac{7x}{2}}}=\frac{\cos \frac{7x}{2}+sin\frac{7x}{2}}{\cos \frac{7x}{2}-sin\frac{7x}{2}}$

$\Rightarrow \sqrt{{\left(\frac{sin\frac{7x}{2}+\cos \frac{7x}{2}}{sin\frac{7x}{2}-\cos \frac{7x}{2}}\right)}^2}=\frac{\cos \frac{7x}{2}+sin\frac{7x}{2}}{\cos \frac{7x}{2}-sin\frac{7x}{2}}$

$\Rightarrow \sqrt{{\left(\frac{\cos \frac{7x}{2}+sin\frac{7x}{2}}{\cos \frac{7x}{2}-sin\frac{7x}{2}}\right)}^2}=\frac{\cos \frac{7x}{2}+sin\frac{7x}{2}}{\cos \frac{7x}{2}-sin\frac{7x}{2}}$

$\Rightarrow \frac{\cos \frac{7x}{2}+sin\frac{7x}{2}}{\cos \frac{7x}{2}-sin\frac{7x}{2}}=\frac{\cos \frac{7x}{2}+sin\frac{7x}{2}}{\cos \frac{7x}{2}-sin\frac{7x}{2}}$

L.H.S.=R.H.S.

Hence this is the answer.