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Question: In ![](https://cdn.pureessence.tech/canvas_122.png?top_left_x=0&top_left_y=1189&width=300&height=46)...

In

A

a2a ^ { 2 }

B

b2b ^ { 2 }

C

c2c ^ { 2 }

D

None of these

Answer

c2c ^ { 2 }

Explanation

Solution

(a2+b2−2ab)cos⁡2C2+(a2+b2+2ab)sin⁡2C2\left( a ^ { 2 } + b ^ { 2 } - 2 a b \right) \cos ^ { 2 } \frac { C } { 2 } + \left( a ^ { 2 } + b ^ { 2 } + 2 a b \right) \sin ^ { 2 } \frac { C } { 2 }

=a2+b2+2ab(sin⁡2C2−cos⁡2C2)= a ^ { 2 } + b ^ { 2 } + 2 a b \left( \sin ^ { 2 } \frac { C } { 2 } - \cos ^ { 2 } \frac { C } { 2 } \right)

=a2+b2−2abcos⁡C=a2+b2−(a2+b2−c2)=c2= a ^ { 2 } + b ^ { 2 } - 2 a b \cos C = a ^ { 2 } + b ^ { 2 } - \left( a ^ { 2 } + b ^ { 2 } - c ^ { 2 } \right) = c ^ { 2 } .