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If $(\sin x+\operatorname{cosec} x)^{2}+(\cos x+\sec x)^{2}=k+\tan ^{2} x+\cot ^{2} x$
then what is the value of $k ?$
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then what is the value of $k ?$
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Verified Answer
The correct answer is:
$\underline{7}$
Given, $(\sin x+\operatorname{cosec} x)^{2}+(\cos x+\sec x)^{2}$
$=k+\tan ^{2} x+\cot ^{2} x$
$\Rightarrow \sin ^{2} x+\operatorname{cosec}^{2} x+2 \sin x \operatorname{cosec} x+\cos ^{2} x+\sec ^{2} x$
$+2 \sec x \cos x=k+\tan ^{2} x+\cot ^{2} x$
$\Rightarrow \sin ^{2} x+\operatorname{cosec}^{2} x+2+\cos ^{2} x+\sec ^{2} x+2$
$\quad=k+\tan ^{2} x+\cot ^{2} x$
$(\because \sin x \operatorname{cosec} x=1$ and $\sec x \cos x=1)$
$\Rightarrow 1+\operatorname{cosec}^{2} x-\cot ^{2} x+\sec ^{2} x-\tan ^{2} x+4=k$
$\Rightarrow 1+1+1+4=k \Rightarrow k=7$
$=k+\tan ^{2} x+\cot ^{2} x$
$\Rightarrow \sin ^{2} x+\operatorname{cosec}^{2} x+2 \sin x \operatorname{cosec} x+\cos ^{2} x+\sec ^{2} x$
$+2 \sec x \cos x=k+\tan ^{2} x+\cot ^{2} x$
$\Rightarrow \sin ^{2} x+\operatorname{cosec}^{2} x+2+\cos ^{2} x+\sec ^{2} x+2$
$\quad=k+\tan ^{2} x+\cot ^{2} x$
$(\because \sin x \operatorname{cosec} x=1$ and $\sec x \cos x=1)$
$\Rightarrow 1+\operatorname{cosec}^{2} x-\cot ^{2} x+\sec ^{2} x-\tan ^{2} x+4=k$
$\Rightarrow 1+1+1+4=k \Rightarrow k=7$
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