Question

If $\sin x+\sin y=a$ and $\cos x+\cos y=b$, then ${\tan }^2\left(\begin{array}{c}\frac{x+y}{2}\end{array}\right)+{\tan }^2\left(\begin{array}{c}\frac{x-y}{2}\end{array}\right)$ is equal to

• A. $\frac{a^4+b^4+4b^2}{a^2{b}^2+b^4}$
• B. $\frac{a^4-b^4+4b^2}{a^2{b}^2+b^4}$
• C. $\frac{a^4-b^4+4a^2}{a^2{b}^2+a^4}$
• D. None of the above

Answer 1

Answer: (B)

$\frac{a^4-b^4+4b^2}{a^2{b}^2+b^4}$

We have,

$\sin x+\sin y=a$ ............. $\left(i\right)$

$\cos x+\cos y=b$ ............ $\left(ii\right)$

On squaring eqn. (i) and (ii) and adding them, we get

$1+1+2(\sin x\sin y+\cos x\cos y)=a^2+b^2$

or, $2+2\cos (x-y)=a^2+b^2$

or, $\cos (x-y)=\frac{a^2+b^2-2}{2}$ ............ $\left(iii\right)$

On dividing eqn.(i) by (ii), we get

$\frac{\sin x+\sin y}{\cos x+\cos y}=\frac{a}{b}$

or, $\tan \frac{(x+y)}{2}=\frac{a}{b}$ ........... $\left(iv\right)$

$\therefore {\tan }^2\frac{(x+y)}{2}+{\tan }^2\frac{(x-y)}{2}$

$=\frac{a^2}{b^2}+\frac{1-\cos (x-y)}{1+\cos (x-y)}$

$=\frac{a^2}{b^2}+\frac{1-\frac{a^2+b^2-2}{2}}{1+\frac{a^2+b^2-2}{2}}$

$=\frac{a^2}{b^2}+\frac{4-a^2-b^2}{a^2+b^2}$

$=\frac{a^4+a^2{b}^2+4b^2-a^2{b}^2-b^4}{a^2{b}^2+b^4}$

$=\frac{a^4-b^4+4b^2}{a^2{b}^2+b^4}$

Hence, B is the correct option.