Question

Assertion :If $2\cos \theta +\sin \theta =1(\theta \ne \frac{\pi }{2})$ then the value of $7\cos \theta +6\sin \theta$ is $2$ Reason: If $\cos 2\theta -\sin \theta =\frac{1}{2},0<\theta <\frac{\pi }{2}$, then $\sin \theta +\cos 6\theta =0$

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Both Assertion and Reason are incorrect

Answer 1

The correct option is B Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion

Assertion: $2\cos \theta +\sin \theta =1\Rightarrow \frac{2(1-{\tan }^2\left(\frac{\theta }{2}\right))}{1+{\tan }^2\left(\frac{\theta }{2}\right)}+\frac{2\tan \left(\frac{\theta }{2}\right)}{1+{\tan }^2\left(\frac{\theta }{2}\right)}=1$

$\Rightarrow 3{\tan }^2\left(\frac{\theta }{2}\right)-2\tan \left(\frac{\theta }{2}\right)-1=0$

$\Rightarrow \tan \left(\frac{\theta }{2}\right)=-\frac{1}{3}$ as $\theta \ne \frac{\pi }{2}$

Now $7\cos \theta +6\sin \theta =\frac{7(1-{\tan }^2\left(\frac{\theta }{2}\right))}{1+{\tan }^2\left(\frac{\theta }{2}\right)}+\frac{6\times 2\tan \left(\frac{\theta }{2}\right)}{1+{\tan }^2\left(\frac{\theta }{2}\right)}$

$=\frac{7-7{\tan }^2\left(\frac{\theta }{2}\right)+12\tan \left(\frac{\theta }{2}\right)}{1+{\tan }^2\left(\frac{\theta }{2}\right)}=\frac{7-7\times \frac{1}{9}+12(-\frac{1}{3})}{1+\frac{1}{9}}=2$

Reason:

$\cos 2\theta -\sin \theta =\frac{1}{2}\Rightarrow 2(1-2{\sin }^2\theta )-\sin \theta =1$

$\Rightarrow 4{\sin }^2\theta +2\sin \theta -1=0\Rightarrow \sin \theta =\frac{-2\pm \sqrt{4+16}}{8}=\frac{-1\pm \sqrt{5}}{4}$

$\Rightarrow \sin \theta =\frac{\sqrt{5}-1}{4}\Rightarrow \theta ={18}^0$

$\Rightarrow \cos 6\theta =\cos {108}^0=\cos ({90}^0+{18}^0)=-\sin {18}^0$

$\Rightarrow \sin \theta +\cos 6\theta =0$

So assertion and reason both are true but reason does not lead assertion.