Question

If $\frac{\cos (\theta -\alpha )}{\sin (\theta +\alpha )}=\frac{m+1}{m-1}$, then m is equal to

• A. $\tan (\frac{\pi }{4}-\theta )\tan (\frac{\pi }{4}-\alpha )$
• B. $\tan (\frac{\pi }{4}-\theta )\tan (\frac{\pi }{4}+\alpha )$
• C. $\tan (\frac{\pi }{4}+\theta )\tan (\frac{\pi }{4}+\alpha )$
• D. $tan(\frac{\pi }{4}+\theta )tan(\frac{\pi }{4}-\alpha )$

Answer 1

Answer: (C)

$\tan (\frac{\pi }{4}+\theta )\tan (\frac{\pi }{4}+\alpha )$

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

Applying componendo and dividendo

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

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

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

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

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

$=\frac{1+\tan \theta }{1-\tan \theta }\cdot \frac{1+\tan \alpha }{1-\tan \alpha }=\tan (\frac{\pi }{4}+\theta )\cdot \tan (\frac{\pi }{4}+\alpha )$