(B) We have,$(a-b)^2 \cos^2 \frac{C}{2} + (a+b)^2 \sin^2 \frac{C}{2}$ $= (a^2 + b^2 - 2ab) \cos^2 \frac{C}{2} + (a^2 + b^2 + 2ab) \sin^2 \frac{C}{2}$

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(B) We have,$(a-b)^2 \cos^2 \frac{C}{2} + (a+b)^2 \sin^2 \frac{C}{2}$ $= (a^2 + b^2 - 2ab) \cos^2 \frac{C}{2} + (a^2 + b^2 + 2ab) \sin^2 \frac{C}{2}$ $= (a^2 + b^2)(\cos^2 \frac{C}{2} + \sin^2 \frac{C}{2}) - 2ab(\cos^2 \frac{C}{2} - \sin^2 \frac{C}{2})$ $= (a^2 + b^2)(1) - 2ab \cos C$ Since $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$,we have $a^2 + b^2 - 2ab \cos C = c^2$ Therefore,the expression is equal to $c^2$.

Class 11 Mathematics Trigonometrical Equations Relation between sides and angles, Solutions of triangles

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In a $\triangle ABC$,the expression $(a-b)^2 \cos^2 \frac{C}{2} + (a+b)^2 \sin^2 \frac{C}{2}$ is equal to:

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AP EAMCET 2004 All AP EAMCET

A
$a^2$
B
$c^2$
C
$b^2$
D
$a^2+b^2$

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Trigonometrical Equations

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Relation between sides and angles, Solutions of triangles

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