Question

If $\frac{\sin (\theta +\alpha )}{\cos (\theta -\alpha )}=\frac{1-m}{1+m}$, then value of $\tan ({45}^{\circ }-\theta )\tan ({45}^{\circ }-\alpha )$ equals

• A. $\frac{1}{1+m}$
• B. $\frac{m+1}{1-m}$
• C. $m$
• D. $\frac{m-1}{m+1}$

Answer 1

The correct option is C $m$

$\frac{\sin (\theta -\alpha )}{\cos (\theta -\alpha )}=\frac{1-m}{1+m}$

Taking componendo and dividendo

$\frac{\sin (\theta +\alpha )+\cos (\theta -\alpha )}{\sin (\theta +\alpha )-\cos (\theta -\cos )}=\frac{1-m+1+m}{1-m-1-m}$

$\Rightarrow \frac{2\sin \theta \cos \alpha }{2\cos \theta \sin \alpha }=-\frac{1}{m}$

$\Rightarrow \frac{\tan \theta }{\tan \alpha }=-\frac{1}{m}$

$\Rightarrow m\tan \theta =-\tan \alpha =\tan (-\alpha )=\tan (90-\alpha )=\cot \alpha$

Again takion componendo and dividendo

$m\left(\frac{1+\tan \theta }{1-\tan \theta }\right)=\left(\frac{1+\cot \alpha }{1-\cot \alpha }\right)$

$\Rightarrow \frac{1}{m}\left(\frac{1-\tan \theta }{1+\tan \theta }\right)=\left(\frac{1-\cot \alpha }{1+\cot \alpha }\right)$

$\Rightarrow \frac{1}{m}\tan (45-\theta )=\cot (45-\alpha )$

$\Rightarrow \tan (45-\theta )\tan (45-\alpha )=m$