If x+ y=a and x+ y=b , then ^2 ((x+y)/(2))+ ^2 ((x-y)/(2))is equal to

Correct answer: B. (a^4-b^4+4 b^2)/(a^2 b^2+b^4) aligned & x+ y=a & 2 ( x+y 2 ) ( x-y 2 )=a....(1) & x+ y=b & 2 ( x+y 2 ) ( x-y 2 )=b....(2) aligned dividing eq (1) \& (2) ( x+y 2 )= a b Squaring eq (1) \& (2) and adding - aligned & 4 ^2((x-y)/(2))=a^2+b^2 & ^2((x-y)/(2))=(4)/(a^2+b^2) aligned again- aligned & ^2( x+y 2 )+ ^2( x-y 2 ) = & ( a b )^2+ ^2( x-y 2 )-1 = & a^2 b^2 + 4 a^2+b^2 -1 = & a^4-b^4+4 b^2 a^2 b^2+b^4 aligned

NDA 2015

Exam: NDA
Year: 2015
Subject: Mathematics
Chapter: Trigonometric Ratios & Identities

NDA 2015 · Mathematics · Trigonometric Ratios & Identities

If x+ y=a and x+ y=b , then ^2 ((x+y)/(2))+ ^2 ((x-y)/(2))is equal...

\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

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 $\frac{a^4-b^4+4 a^2}{a^2 b^2+a^4}$
DNone of the above

Correct answer

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

Step-by-step solution

\begin{aligned} & \sin \mathrm{x}+\sin \mathrm{y}=\mathrm{a} \\ & \Rightarrow \quad 2 \sin \left(\frac{\mathrm{x}+\mathrm{y}}{2}\right) \cos \left(\frac{\mathrm{x}-\mathrm{y}}{2}\right)=\mathrm{a}....(1) \\ & \cos \mathrm{x}+\cos \mathrm{y}=\mathrm{b} \\ & \Rightarrow \quad 2 \cos \left(\frac{\mathrm{x}+\mathrm{y}}{2}\right) \cos \left(\frac{\mathrm{x}-\mathrm{y}}{2}\right)=\mathrm{b}....(2) \end{aligned}

dividing eq (1) \& (2)

\tan \left(\frac{\mathrm{x}+\mathrm{y}}{2}\right)=\frac{\mathrm{a}}{\mathrm{ b}}

Squaring eq (1) \& (2) and adding -

\begin{aligned} & 4 \cos ^2\left(\frac{x-y}{2}\right)=a^2+b^2 \\ & \sec ^2\left(\frac{x-y}{2}\right)=\frac{4}{a^2+b^2} \end{aligned}

again-

\begin{aligned} & \tan ^2\left(\frac{\mathrm{x}+\mathrm{y}}{2}\right)+\tan ^2\left(\frac{\mathrm{x}-\mathrm{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} \end{aligned}

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