If sinA=sinB and cosA=cosB, then

sin $\frac{A - B}{2}$ = 0

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sin $\frac{A + B}{2}$ = 0

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cos $\frac{A - B}{2}$ = 0

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cos (A + B) = 0

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Solution

The correct option is A
sin $\frac{A - B}{2}$ = 0

We have sin A = sin B and cos A = cos B

$2 c o s (\frac{A + B}{2}) s i n (\frac{A - B}{2}) = 0, - 2 s i n (\frac{A + B}{2}) s i n (\frac{A - B}{2}) = 0$

Hence, sin $(\frac{A - B}{2})$ = 0.

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Similar questions

If $sin A = sin B$ and $cos A = cos B$ , then

A. $sin \frac{A - B}{2} = 0$

B. $sin \frac{A + B}{2} = 0$

C. $cos \frac{A - B}{2} = 0$

cos A + B = 0

sin A + cos B = a

sin B + cos A = b

sin (A + B)

$\frac{s i n A - s i n B}{c o s A + c o s B} + \frac{c o s A - c o s B}{s i n A + s i n B} =$

\frac{sin A - sin B}{cos A + cos B} + \frac{cos A - cos B}{sin A + sin B} = 0

Simplify the expression:
$\frac{s i n A - s i n B}{c o s A + c o s B} + \frac{c o s A - c o s B}{s i n A + s i n B}$

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