Question

$\frac{\sin (\theta +\alpha )}{\cos (\theta -\alpha )}=\frac{1-m}{1+m}$, $\tan (\frac{\pi }{4}-\theta )\tan (\frac{\pi }{4}-\alpha )$ =

• A. $\frac{1}{m}$
• B. $m$
• C. $\frac{2}{m}$
• D. $2m$

Answer 1

Answer: (B)

$m$

$\frac{\sin (\theta +\alpha )}{\cos (\theta -\alpha )}=\frac{1-m}{1+m}$
$m\left(\sin \right(\theta +\alpha )+\cos (\theta -\alpha \left)\right)=\cos (\theta -\alpha )-\sin (\theta +\alpha )$
$m=\frac{\cos \theta \cos \alpha +\sin \theta \sin \alpha -\sin \theta \cos \alpha -\sin \alpha \cos \theta }{\sin \theta \cos \alpha +\cos \theta \sin \alpha +\cos \theta \cos \alpha +\sin \theta \sin \alpha }$
$=\frac{\cos \alpha (\cos \theta -\sin \theta )-\sin \alpha (\cos \theta -\sin \theta )}{\cos \alpha (\sin \theta +\cos \theta )+\sin \alpha (\sin \theta +\cos \theta )}$
$=\frac{(\cos \theta -\sin \theta )(\cos \alpha -\sin \alpha )}{(\sin \theta +\cos \theta )(\cos \alpha +\sin \alpha )}$
$=\frac{\sin (\frac{\pi }{4}-\theta )\sin (\frac{\pi }{4}-\alpha )}{\cos (\frac{\pi }{4}-\theta )\cos (\frac{\pi }{4}-\alpha )}$
$\Rightarrow m=\tan (\frac{\pi }{4}-\theta )\tan (\frac{\pi }{4}-\alpha )$