Question

(i) If $\sin (\theta +\alpha )=\cos (\theta +\alpha )$ then express $\tan \theta$ in terms of $\alpha$.
(ii) Find the value of $\tan (\pi /4+\theta ).\tan (\pi /4-\theta )$

Answer 1

(i) $\sin (\theta +\alpha )=\cos (\theta +\alpha )$

$\sin \theta .\cos \alpha +\cos \theta .\sin \alpha =\cos \theta .\cos \alpha -\sin \theta .\sin \alpha$

Diving the complete equation by $\cos \theta$, and writing $\frac{\sin \theta }{\cos \theta }=\tan \theta$, we get

$\tan \theta .\cos \alpha +\sin \alpha =\cos \alpha -\tan \theta .\sin \alpha$

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

(ii)$\tan (\pi /4+\theta ).\tan (\pi /4-\theta )$

$\frac{\tan \left(\pi /4\right)+\tan \theta }{1-\tan \left(\pi /4\right).\tan \theta }.\frac{\tan \left(\pi /4\right)-\tan \theta }{1+\tan \left(\pi /4\right).\tan \theta }$

$\frac{1+\tan \theta }{1-\tan \theta }.\frac{1-\tan \theta }{1+\tan \theta }$

$=1$