Search any question & find its solution
Question:
Answered & Verified by Expert
If $\sin x+\sin y=a$ and $\cos x+\cos y=b$, then
$\tan ^{2}\left(\frac{x+y}{2}\right)+\tan ^{2}\left(\frac{x-y}{2}\right)$ is equal to
Options:
$\tan ^{2}\left(\frac{x+y}{2}\right)+\tan ^{2}\left(\frac{x-y}{2}\right)$ is equal to
Solution:
1090 Upvotes
Verified Answer
The correct answer is:
$\frac{a^{4}-b^{4}+4 b^{2}}{a^{2} b^{2}+b^{4}}$
$\sin x+\sin y=a$
$\Rightarrow 2 \sin \left(\frac{\mathrm{x}+\mathrm{y}}{2}\right) \cos \left(\frac{\mathrm{x}-\mathrm{y}}{2}\right)=\mathrm{a}$
$\cos x+\cos y=b$
$\Rightarrow \quad 2 \cos \left(\frac{\mathrm{x}+\mathrm{y}}{2}\right) \cos \left(\frac{\mathrm{x}-\mathrm{y}}{2}\right)=\mathrm{b}$
dividing eq $(1) \&(2)$ $\tan \left(\frac{x+y}{2}\right)=\frac{a}{b}$
Squaring eq (1) \& (2) and adding $4 \cos ^{2}\left(\frac{\mathrm{x}-\mathrm{y}}{2}\right)=\mathrm{a}^{2}+\mathrm{b}^{2}$
$\sec ^{2}\left(\frac{x-y}{2}\right)=\frac{4}{a^{2}+b^{2}}$
again, $\tan ^{2}\left(\frac{x+y}{2}\right)+\tan ^{2}\left(\frac{x-y}{2}\right)$
$=\left(\frac{\mathrm{a}}{\mathrm{b}}\right)^{2}+\sec ^{2}\left(\frac{\mathrm{x}-\mathrm{y}}{2}\right)-1$
$=\frac{\mathrm{a}^{2}}{\mathrm{~b}^{2}}+\frac{4}{\mathrm{a}^{2}+\mathrm{b}^{2}}-1=\frac{\mathrm{a}^{4}-\mathrm{b}^{4}+4 \mathrm{~b}^{2}}{\mathrm{a}^{2} \mathrm{~b}^{2}+\mathrm{b}^{4}}$
$\Rightarrow 2 \sin \left(\frac{\mathrm{x}+\mathrm{y}}{2}\right) \cos \left(\frac{\mathrm{x}-\mathrm{y}}{2}\right)=\mathrm{a}$
$\cos x+\cos y=b$
$\Rightarrow \quad 2 \cos \left(\frac{\mathrm{x}+\mathrm{y}}{2}\right) \cos \left(\frac{\mathrm{x}-\mathrm{y}}{2}\right)=\mathrm{b}$
dividing eq $(1) \&(2)$ $\tan \left(\frac{x+y}{2}\right)=\frac{a}{b}$
Squaring eq (1) \& (2) and adding $4 \cos ^{2}\left(\frac{\mathrm{x}-\mathrm{y}}{2}\right)=\mathrm{a}^{2}+\mathrm{b}^{2}$
$\sec ^{2}\left(\frac{x-y}{2}\right)=\frac{4}{a^{2}+b^{2}}$
again, $\tan ^{2}\left(\frac{x+y}{2}\right)+\tan ^{2}\left(\frac{x-y}{2}\right)$
$=\left(\frac{\mathrm{a}}{\mathrm{b}}\right)^{2}+\sec ^{2}\left(\frac{\mathrm{x}-\mathrm{y}}{2}\right)-1$
$=\frac{\mathrm{a}^{2}}{\mathrm{~b}^{2}}+\frac{4}{\mathrm{a}^{2}+\mathrm{b}^{2}}-1=\frac{\mathrm{a}^{4}-\mathrm{b}^{4}+4 \mathrm{~b}^{2}}{\mathrm{a}^{2} \mathrm{~b}^{2}+\mathrm{b}^{4}}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.