Question

Assertion :In a $△ABC,$ if
${\cos }^2\frac{A}{2}+{\cos }^2\frac{B}{2}+{\cos }^2\frac{C}{2}=y(x^2+\frac{1}{x^2})$
then the maximum value of $y$ is $\frac{9}{8}$ Reason: In a $△ABC,\sin \frac{A}{2}.\sin \frac{B}{2}.\sin \frac{C}{2}\le \frac{1}{8}$

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

Answer 1

The correct option is A Both Assertion and Reason are individually true and Reason is the correct explanation of Assertion.

${\cos }^2\frac{A}{2}+{\cos }^2\frac{B}{2}+{\cos }^2\frac{C}{2}$
$=\frac{1+\cos A}{2}+\frac{1+\cos B}{2}+\frac{1+\cos C}{2}$
$=\frac{1}{2}[3+\cos A+\cos B+\cos C]$
$=\frac{1}{2}[3+2\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)+1-2{\sin }^2\frac{C}{2}]$
$=\frac{1}{2}[4+2\cos (\frac{\pi }{2}-\frac{C}{2})\cos \left(\frac{A-B}{2}\right)-2{\sin }^2\frac{C}{2}]$
$=\frac{1}{2}[4+2\sin \frac{C}{2}\cos \left(\frac{A-B}{2}\right)-2{\sin }^2\frac{C}{2}]$
$=[2+\sin \frac{C}{2}\cos \left(\frac{A-B}{2}\right)-{\sin }^2\frac{C}{2}]$
$=2+\sin \frac{C}{2}[\cos \left(\frac{A-B}{2}\right)-\sin \frac{C}{2}]$
$=2+\sin \frac{C}{2}[\cos \left(\frac{A-B}{2}\right)-\sin (\frac{\pi }{2}-\frac{A+B}{2})]$
$=2+\sin \frac{C}{2}[\cos \left(\frac{A-B}{2}\right)-\cos \left(\frac{A+B}{2}\right)]$
$=2+2\sin \frac{C}{2}\sin \frac{A}{2}\sin \frac{B}{2}$ by transformation angle formula.
$=2(1+\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2})$
Now $2(1+\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2})=y(x^2+\frac{1}{x^2})$(given)
$\ge 2y$ ($\because$A.M$\ge$G.M)
$\therefore y\le \frac{1}{2}\times 2(1+\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2})$
$\Rightarrow y\le (1+\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2})\le 1+\frac{1}{8}$
i.e., $y\le \frac{9}{8}$
$\therefore$ Maximum value of $y$ is $\frac{9}{8}$