Question

The possible value(s) of '$\theta$' satisfying the equation ${\sin }^2\theta \tan \theta +{\cos }^2\theta \cot \theta -\sin 2\theta =1+\tan \theta +\cot \theta$ , where $\theta \in [0,\pi ]$ is/are

• A. $\frac{\pi }{4}$
• B. $\pi$
• C. $\frac{7\pi }{12}$
• D. None of these

Answer 1

The correct option is C $\frac{7\pi }{12}$

${\sin }^2\theta \tan \theta +{\cos }^2\theta \cot \theta -\sin 2\theta =1+\cot \theta +\tan \theta$ ...(1)
$\theta \ne n\pi$ because equation (1) is not defined
$\Rightarrow \frac{{\sin }^4\theta +{\cos }^4\theta }{\sin \theta \cos \theta }-\sin 2\theta =1+\frac{{\sin }^2\theta +{\cos }^2\theta }{\sin \theta \cos \theta }\Rightarrow \frac{{\left(\frac{1-\cos 2\theta }{2}\right)}^2+{\left(\frac{1+\cos 2\theta }{2}\right)}^2}{\sin \theta \cos \theta }-\sin 2\theta =1+\frac{2}{2\sin \theta \cos \theta }\Rightarrow \frac{1+{\cos }^{2}2\theta }{\sin 2\theta }-\sin 2\theta =1+\frac{2}{\sin 2\theta }\Rightarrow 1+1-{\sin }^{2}2\theta -{\sin }^{2}2\theta =\sin 2\theta +2\Rightarrow -2{\sin }^{2}2\theta =\sin 2\theta \because \theta \ne n\pi \therefore \sin 2\theta =\frac{-1}{2}\Rightarrow \theta =\frac{7\pi }{12},\frac{5\pi }{6}$
Ans: C