Question

Match $\text{List I}$ with the $\text{List II}$ and select the correct answer using the code given below the lists :

$\begin{array}{cccc}& \text{List I}& & \text{List II}\\ \text{(A)}& 4({\sin }^6\frac{\pi }{16}+{\sin }^6\frac{3\pi }{16}+{\sin }^6\frac{5\pi }{16}+{\sin }^6\frac{7\pi }{16})\text{equals}& \text{(P)}& 2\\ \text{(B)}& \text{If}\alpha =\frac{2\pi }{7},\text{then the value of}5+\sec \alpha +\sec 2\alpha +\sec 4\alpha \text{is}& \text{(Q)}& 4\\ \text{(C)}& {\tan }^2\frac{\pi }{16}+{\tan }^2\frac{3\pi }{16}+{\tan }^2\frac{5\pi }{16}+{\tan }^2\frac{7\pi }{16}-24\text{equals}& \text{(R)}& 5\\ \text{(D)}& \text{The maximum value of}& \text{(S)}& 1\\ & 2(7\cos \theta +24\sin \theta )\times (7\sin \theta -24\cos \theta )-623\text{for}\theta \in R\text{is}& & \end{array}$

Which of the following is the only CORRECT combination?

• A. $\left(C\right)\to \left(P\right),\left(D\right)\to \left(R\right)$
• B. $\left(C\right)\to \left(Q\right),\left(D\right)\to \left(S\right)$
• C. $\left(C\right)\to \left(2\right),\left(D\right)\to \left(2\right)$
• D. $\left(C\right)\to \left(Q\right),\left(D\right)\to \left(P\right)$

Answer 1

The correct option is D $\left(C\right)\to \left(Q\right),\left(D\right)\to \left(P\right)$

Let $\frac{\pi }{16}=\theta$
${\tan }^2\theta +{\tan }^{2}3\theta +{\tan }^{2}5\theta +{\tan }^{2}7\theta$
$=({\tan }^2\theta +{\cot }^2\theta )+({\tan }^{2}3\theta +{\cot }^{2}3\theta )$
$[\because \tan 7\theta =\tan (8\theta -\theta )=\cot \theta$ and $\tan 5\theta =\tan (8\theta -3\theta )=\cot 3\theta ]$
$=(\cot \theta -\tan \theta {)}^2+(\cot 3\theta -\tan 3\theta {)}^2+4$
$=4[{\cot }^{2}2\theta +{\cot }^{2}6\theta ]+4$
$=4[{\cot }^{2}2\theta +{\tan }^{2}2\theta ]+4$
$=4\left[\right(\cot 2\theta -\tan 2\theta {)}^2+2]+4$
$=4(\cot 2\theta -\tan 2\theta {)}^2+12$
$=4\cdot 4{\cot }^{2}4\theta +12=16\times 1^2+12=28.$

$y=(7\cos \theta +24\sin \theta )\times (7\sin \theta -24\cos \theta )$
$r\cos \varphi =7;r\sin \varphi =24$
$r^2=25;\tan \varphi =\frac{24}{7}$
$y=r\cos (\theta -\varphi )\cdot r\sin (\theta -\varphi )$
$=\frac{r^2}{2}\cdot 2\sin (\theta -\varphi )\cos (\theta -\varphi )$
$=\frac{r^2}{2}\cdot \left(\sin 2\right(\theta -\varphi \left)\right)$
$\Rightarrow y_{max}=\frac{{25}^2}{2}=\frac{625}{2}$
Now , $2\times \frac{625}{2}-623=2$