In a triangle $\mathrm{ABC}$, with usual notations, if $\mathrm{c}=4$, then value of $(a-b)^2 \cos ^2…
In a triangle $\mathrm{ABC}$, with usual notations, if $\mathrm{c}=4$, then value of $(a-b)^2 \cos ^2 \frac{C}{2}+(a+b)^2 \sin ^2 \frac{C}{2}$ is
- $4$
- $16$
- $9$
- $2$
Solution
$\begin{aligned} & (a-b)^2 \cos ^2 \frac{C}{2}+(a+b)^2 \sin ^2 \frac{C}{2} \\ & =\left(a^2-2 a b+b^2\right) \cos ^2 \frac{C}{2}+\left(a^2+2 a b+b^2\right) \sin ^2 \frac{C}{2} \\ & =\left(a^2+b^2\right)\left(\cos ^2 \frac{C}{2}+\sin ^2 \frac{C}{2}\right) \\ & \quad-2 a b \cos ^2 \frac{C}{2}+2 a b \sin ^2 \frac{C}{2} \\ & =a^2+b^2-2 a b\left(\cos ^2 \frac{C}{2}-\sin ^2 \frac{C}{2}\right) \\ & =a^2+b^2-2 a b \cdot \cos C \\ & =a^2+b^2-\left(a^2+b^2-c^2\right) \quad \ldots[\text { By Cosine rule }] \\ & =c^2=4^2=16\end{aligned}$
Asked in: MHT CET 2023 (10 May Shift 2)
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