Question

If $\sin (\theta +\alpha )=\cos (\theta +\alpha )$, then which of the following is/are correct?
where $\theta ,\alpha \in (0,\frac{\pi }{2})-\left\{\frac{\pi }{4}\right\}$

• A. $\tan \alpha =\frac{1-\tan \theta }{1+\tan \theta }$
• B. $\tan \alpha =\frac{1+\tan \theta }{1-\tan \theta }$
• C. $\tan \theta =\frac{1+\tan \alpha }{1-\tan \alpha }$
• D. $\tan \theta =\frac{1-\tan \alpha }{1+\tan \alpha }$

Answer 1

Answer: (A), (D)

$\tan \theta =\frac{1-\tan \alpha }{1+\tan \alpha }$

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

Now,
$\tan \theta =\frac{1-\tan \alpha }{1+\tan \alpha }\Rightarrow \tan \theta +\tan \alpha \tan \theta =1-\tan \alpha \Rightarrow \tan \alpha =\frac{1-\tan \theta }{1+\tan \theta }$