In a Δ ABC, {²≤ft(B-C/2)}{(b+c)² + {²≤ft(B-C/2)}{(b-c)² =

Correct Answer is : 1a21/a²a21

Correct Answer is : 1a21/a²a21

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TS EAMCET 5-Aug-2021 Shift 2 Question Paper

Section: Mathematics

Question : 30 of 160

Marks: +1, -0

In a Δ ABC,

\frac{\cos^2\left(\frac{B-C}{2}\right)}{(b+c)^2} + \frac{\sin^2\left(\frac{B-C}{2}\right)}{(b-c)^2} =

Solution: 👈: Video Solution

\; \frac{\cos^{2}\left(\frac{B-C}{2}\right)}{(b+c)^{2}} + \; \frac{\sin^{2}\left(\frac{B-C}{2}\right)}{(b-c)^{2}}

\begin{array}{l} \text{ by using sine theorem } \\ \frac{\sin B}{b} = \frac{\sin C}{c} = \frac{1}{k} \text{ (let) } \end{array}

(b+c)^2 = k^2 (\sin B + \sin C)^2

= 4 k^2 \sin^2\left(\frac{B+C}{2}\right) \cos^2\left(\frac{B-C}{2}\right)

\Rightarrow 4 k^2 \sin^2\left(\frac{180-A}{2}\right) \cos^2\left(\frac{B-C}{2}\right)

\Rightarrow \; \frac{(b+c)^2}{\cos^2\left(\frac{B-C}{2}\right)} = 4 k^2 \cos^2\left(\frac{A}{2}\right)

\Rightarrow \; \frac{1}{4 k^2 \cos^2\left(\frac{A}{2}\right)} + \; \frac{1}{4 k^2 \sin^2\left(\frac{A}{2}\right)}

\Rightarrow \; \frac{\cos^2\left(\frac{A}{2}\right) + \sin^2\left(\frac{A}{2}\right)}{4 k^2 \cos^2\left(\frac{A}{2}\right) \sin^2\left(\frac{A}{2}\right)}

\Rightarrow \; \frac{1}{\left(2 k \sin \frac{A}{2} \cos \frac{A}{2}\right)^2} \Rightarrow \; \frac{1}{k^2 \sin^2 A} = \; \frac{1}{a^2}

\; a = k \sin A

\text{SO}, \; \frac{1}{a^2}

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