In a triangle ABC , with usual notations, if c=4 , then value of (a-b)^2 ^2 (C)/(2)+(a+b)^2 ^2 (C)/(2) is

Correct answer: B. 16 aligned & (a-b)^2 ^2 (C)/(2)+(a+b)^2 ^2 (C)/(2) & =(a^2-2 a b+b^2) ^2 (C)/(2)+(a^2+2 a b+b^2) ^2 (C)/(2) & =(a^2+b^2)( ^2 (C)/(2)+ ^2 (C)/(2)) & -2 a b ^2 (C)/(2)+2 a b ^2 (C)/(2) & =a^2+b^2-2 a b( ^2 (C)/(2)- ^2 (C)/(2)) & =a^2+b^2-2 a b · C & =a^2+b^2-(a^2+b^2-c^2) [ By Cosine rule ] & =c^2=4^2=16 aligned

JEE Main

Exam: JEE Main
Subject: Mathematics
Chapter: Trigonometric Ratios & Identities

JEE Main · Mathematics · Trigonometric Ratios & Identities

In a triangle ABC , with usual notations, if c=4 , then value of...

In a triangle

\mathrm{ABC}

, with usual notations, if

\mathrm{c}=4

, then value of

(a-b)^2 \cos ^2 \frac{C}{2}+(a+b)^2 \sin ^2 \frac{C}{2}

Options

A $4$
B $16$
C $9$
D $2$

Correct answer

B. $16$

Step-by-step solution

\begin{aligned} & (a-b)^2 \cos ^2 \frac{C}{2}+(a+b)^2 \sin ^2 \frac{C}{2} \\ & =\left(a^2-2 a b+b^2\right) \cos ^2 \frac{C}{2}+\left(a^2+2 a b+b^2\right) \sin ^2 \frac{C}{2} \\ & =\left(a^2+b^2\right)\left(\cos ^2 \frac{C}{2}+\sin ^2 \frac{C}{2}\right) \\ & \quad-2 a b \cos ^2 \frac{C}{2}+2 a b \sin ^2 \frac{C}{2} \\ & =a^2+b^2-2 a b\left(\cos ^2 \frac{C}{2}-\sin ^2 \frac{C}{2}\right) \\ & =a^2+b^2-2 a b \cdot \cos C \\ & =a^2+b^2-\left(a^2+b^2-c^2\right) \quad \ldots[\text { By Cosine rule }] \\ & =c^2=4^2=16\end{aligned}

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