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. Assertion is incorrect but Reason is correct

Answer 1

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

Statement 1: is true as
$2\cos \theta +\sin \theta =1\Rightarrow \frac{2(1+{\tan }^2\left(\frac{\theta }{2}\right))}{1+{\tan }^2\left(\frac{\theta }{2}\right)}+\frac{2\tan \left(\frac{\theta }{2}\right)}{1+{\tan }^2\left(\frac{\theta }{2}\right)}=1\Rightarrow 3{\tan }^2\left(\frac{\theta }{2}\right)-2\tan \left(\frac{\theta }{2}\right)-1=0\Rightarrow \tan \left(\frac{\theta }{2}\right)=-\frac{1}{3}\theta \ne \frac{\pi }{2}$
Now,
$7\cos \theta +6\sin \theta =\frac{7(1-{\tan }^2\left(\frac{\theta }{2}\right))}{1+{\tan }^2\left(\frac{\theta }{2}\right)}+\frac{6\times 2\tan \left(\frac{\theta }{2}\right)}{1+{\tan }^2\left(\frac{\theta }{2}\right)}=\frac{7-7{\tan }^2\left(\frac{\theta }{2}\right)+12\tan \left(\frac{\theta }{2}\right)}{1+{\tan }^2\left(\frac{\theta }{2}\right)}=\frac{7-7\times \frac{1}{9}+12(-\frac{1}{3})}{1+\frac{1}{9}}=2$

Statement 2: is also true
$\cos 2\theta -\sin \theta =\frac{1}{2}\Rightarrow 2(1-2{\sin }^2\theta )-\sin \theta =1\Rightarrow 4{\sin }^2\theta +2\sin \theta -1=0\Rightarrow \sin \theta =\frac{-2\pm \sqrt{4+16}}{8}=\frac{-1\pm \sqrt{5}}{4}\Rightarrow \sin \theta =\frac{\sqrt{5}-1}{4}\Rightarrow \theta ={18}^o\Rightarrow \cos 6\theta =\cos {108}^o=\cos ({90}^o+{18}^o)=-\sin {18}^o$
Therefore, $\sin \theta +\cos 6\theta =0$
Hence, both statements is correct but statement 2 is not correct explanation of statement 1