Question

If $\cos x-\sin \alpha \cot \beta \sin x=\cos \alpha$, then $\tan \frac{x}{2}$ is equal to-

• A. $\cot \frac{\alpha }{2}\tan \frac{\beta }{2}$
• B. $-\cot \frac{\beta }{2}\tan \frac{\alpha }{2}$
• C. $\tan \frac{\alpha }{2}\tan \frac{\beta }{2}$
• D. $\cot \frac{\alpha }{2}\cot \frac{\beta }{2}$.

Answer 1

Answer: (B), (C)

The correct options are
B $-\cot \frac{\beta }{2}\tan \frac{\alpha }{2}$
C $\tan \frac{\alpha }{2}\tan \frac{\beta }{2}$

$\cos x-\sin \alpha \cot \beta \sin x=\cos \alpha$

or,$\frac{1-{\tan }^2\frac{x}{2}}{1+{\tan }^2\frac{x}{2}}-\sin \alpha \cot \beta \frac{2\tan \frac{x}{2}}{1+{\tan }^2\frac{x}{2}}=\cos \alpha$

or, $1-{\tan }^2\frac{x}{2}-2\tan \frac{x}{2}\cdot \sin \alpha \cot \beta =\cos \alpha (1+{\tan }^2\frac{x}{2})$

or,${\tan }^2\frac{x}{2}(\cos \alpha +1)+2\tan \frac{x}{2}\sin \alpha \cot \beta +\cos \alpha -1=0$

or, $\tan \frac{x}{2}=-2\sin \alpha \cot \beta \pm \sqrt{4{\sin }^2\alpha \times {\cot }^2\beta -4\times (1+\cos \alpha )(1-\cos \alpha )}$

or, $\tan \frac{x}{2}=\frac{-2\sin \alpha +\cot \beta \pm \sqrt{4{\sin }^2\alpha {\cot }^2\beta +4{\sin }^2\alpha }}{2(1+\cos \alpha )}$

or, $\tan \frac{x}{2}=\frac{-2\sin \alpha \cot \beta \pm 2\sin \alpha \csc \beta }{2(1+\cos \alpha )}$

or, $\tan \frac{x}{2}=\frac{-\sin \alpha \cot \beta \pm 2\sin \alpha \csc \beta }{1+\cos \alpha }$

or, $\tan \frac{x}{2}=\frac{-\sin \alpha \cot \beta +\sin \alpha \csc \beta }{1+\cos \alpha }$

$=\frac{\sin \alpha (\csc \beta -\cot \beta )}{2{\cos }^2\frac{\alpha }{2}}$

$=\tan \frac{\alpha }{2}\cdot \tan \frac{\beta }{2}$

or, $\tan \frac{x}{2}=\frac{-(\sin \alpha \cot \beta +\sin \alpha \csc \beta )}{1+\cos \alpha }$

$=\frac{-2\sin \frac{\alpha }{2}\cdot \cos \frac{\alpha }{2}}{2{\cos }^2\frac{\alpha }{2}}\times \frac{1+\cos \beta }{\sin \beta }$

$=-\tan \frac{\alpha }{2}\cot \frac{\beta }{2}$