Question
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
A.
$$\frac{{{a^4} + {b^4} + 4{b^2}}}{{{a^2}{b^2} + {b^4}}}$$
B.
$$\frac{{{a^4} - {b^4} + 4{b^2}}}{{{a^2}{b^2} + {b^4}}}$$
C.
$$\frac{{{a^4} - {b^4} + 4{a^2}}}{{{a^2}{b^2} + {a^4}}}$$
D.
None of the above
Answer :
$$\frac{{{a^4} - {b^4} + 4{b^2}}}{{{a^2}{b^2} + {b^4}}}$$
Solution :
$$\eqalign{
& \sin x + \sin y = a \cr
& \Rightarrow 2\sin \left( {\frac{{x + y}}{2}} \right)\cos \left( {\frac{{x - y}}{2}} \right) = a\,\,\,.....\left( 1 \right) \cr
& \cos x + \cos y = b \cr
& \Rightarrow 2\cos \left( {\frac{{x + y}}{2}} \right)\cos \left( {\frac{{x - y}}{2}} \right) = b\,\,\,.....\left( 2 \right) \cr} $$
dividing eq $$\left( 1 \right)\,\,\& \,\,\left( 2 \right)$$
$$\tan \left( {\frac{{x + y}}{2}} \right) = \frac{a}{b}$$
Squaring of eq (1) $$\&$$ (2) and adding -
$$\eqalign{
& 4\,{\cos ^2}\left( {\frac{{x - y}}{2}} \right) = {a^2} + {b^2} \cr
& {\sec ^2}\left( {\frac{{x - y}}{2}} \right) = \frac{4}{{{a^2} + {b^2}}}\, \cr
& {\text{again}} , \cr
& {\tan ^2}\left( {\frac{{x + y}}{2}} \right) + {\tan ^2}\left( {\frac{{x - y}}{2}} \right) \cr
& = {\left( {\frac{a}{b}} \right)^2} + {\sec ^2}\left( {\frac{{x - y}}{2}} \right) - 1 \cr
& = \frac{{{a^2}}}{{{b^2}}} + \frac{4}{{{a^2} + {b^2}}} - 1 \cr
& = \frac{{{a^4} - {b^4} + 4{b^2}}}{{{a^2}{b^2} + {b^4}}} \cr} $$