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Question

If A, B and C are interior angles of ΔABC, then Prove that : sin(A+C)2=cosB2.

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Solution

In ΔABC
We know that, Sum of angles of a triangle is 180
A+B+C=180
A+C=180B
Dividing by 2 on both sides
A+C2=180B2
A+C2=90B2
sin (A+C2)=sin(90B2)
sin (A+C2)=cosB2
sin (A+C2)=cosB2
Hence Proved.

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