Question

{(b-c)^2}×{cos^2(A/2)}+{(b+c)^2}×{sin^2(A/2)}​

Answer 1

Answer:

$a^{2}$

Step-by-step explanation:

Using the formulae $(x^{2} - y^{2} ) = x^{2} + y^{2} - 2xy\\(x^{2} + y^{2} ) = x^{2} + y^{2} + 2xy$

We get,

$(c^{2} + b^{2} -2bc) x^{2} \frac{A}{2} + (c^{2} + b^{2} +2bc) x^{2} \frac{A}{2}\\\\=> c^{2} cos^{2} \frac{A}{2} + b^{2} cos^{2} \frac{A}{2} - 2bc cos^{2} \frac{A}{2} + c^{2} sin^{2} \frac{A}{2} + b^{2} sin^{2} \frac{A}{2} + 2bc sin^{2} \frac{A}{2}\\\\=> c^{2}( cos^{2} \frac{A}{2}+ sin^{2} \frac{A}{2}) + b^{2}( cos^{2} \frac{A}{2}+ sin^{2} \frac{A}{2}) -2bc ( cos^{2} \frac{A}{2}- sin^{2} \frac{A}{2})\\\\=> c^{2}+ b^{2}-2bc(\frac{b^{2}+c^{2}-a^{2}}{2bc})$

since, $cos^{2}$∅ - $sin^{2}$∅ == cos2∅

here ∅=A/2

∴ cos2∅ = cos A

so we get,

$=> c^{2} +b^{2} -b^{2} -c^{2} +a^{2} \\=> a^{2}$

hope it helps!!

it took some time to answer it in the proper format

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