Question

If $\frac{m\tan (\alpha -\theta )}{{\cos }^2\theta }=\frac{n\tan \theta }{{\cos }^2(\alpha -\theta )}.$ Find the vale of $\theta$.

• A. $\theta =\frac{1}{2}[\alpha +{\tan }^{-1}\left(\frac{n+m}{n-m}\tan \alpha \right)]$
• B. $\theta =\frac{1}{2}[\alpha -{\tan }^{-1}\left(\frac{n+m}{n-m}\tan \alpha \right)]$
• C. $\theta =\frac{1}{2}[\alpha -{\tan }^{-1}\left(\frac{n-m}{n+m}\tan \alpha \right)]$
• D. $\theta =\frac{1}{2}[\alpha +{\tan }^{-1}\left(\frac{n-m}{n+m}\tan \alpha \right)]$

Answer 1

The correct option is C $\theta =\frac{1}{2}[\alpha -{\tan }^{-1}\left(\frac{n-m}{n+m}\tan \alpha \right)]$

Upon simplification, we get
$\frac{m.sin(\alpha -\theta )}{cos(\alpha -\theta ).co{s}^2\theta }=\frac{n.sin\theta }{cos\theta .co{s}^2(\alpha -\theta )}$
$\frac{m}{n}=\frac{sin\theta .cos\theta }{sin(\alpha -\theta ).cos(\alpha -\theta )}$
$\frac{m}{n}=\frac{sin2\theta }{sin2(\alpha -\theta )}$
$\frac{m-n}{m+n}=\frac{sin2\theta -sin\left(2\right(\alpha -\theta \left)\right)}{sin2\theta +sin\left(2\right(\alpha -\theta \left)\right)}$
$\frac{m-n}{m+n}=\frac{2cos\alpha .sin(2\theta -\alpha )}{2sin\alpha .cos(2\theta -\alpha )}$
$\frac{m-n}{m+n}=cot\alpha .tan(2\theta -\alpha )$
$tan\left(\alpha \right)\left(\frac{m-n}{m+n}\right))=tan(2\theta -\alpha )$
$ta{n}^{-1}\left(tan\right(\alpha \left)\right(\frac{m-n}{m+n}\left)\right))=2\theta -\alpha$
$\alpha +ta{n}^{-1}\left(tan\right(\alpha \left)\right(\frac{m-n}{m+n}\left)\right)=2\theta$
$\theta =\frac{1}{2}[\alpha +ta{n}^{-1}(tan\left(\alpha \right)\left(\frac{m-n}{m+n}\right)\left)\right]$
$=\frac{1}{2}[\alpha +ta{n}^{-1}(tan\left(\alpha \right)(-\frac{n-m}{m+n})\left)\right]$
$=\frac{1}{2}[\alpha +ta{n}^{-1}(-tan\left(\alpha \right)\left(\frac{n-m}{m+n}\right)\left)\right]$
$=\frac{1}{2}[\alpha -ta{n}^{-1}(tan\left(\alpha \right)\left(\frac{n-m}{m+n}\right)\left)\right]$