Find sin2A-sin2Bsin Acos A-sin Bcos B-

The correct option is D tan (A+B) sin^ 2 A sin^ 2 B sin Acos A sin Bcos B = 1 cos 2 A 2 1 cos 2 B 2 #160; 1 2 (sin 2 A sin 2 B) = 1 ...

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Standard XII

Mathematics

Basic Trigonometric Identities

Find sin2A-...

Question

Find $\frac{{sin}^{2} A - {sin}^{2} B}{sin A cos A - sin B cos B} = ?$ .

cos (A + B)

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tan (A - B)

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cot (A + B)

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tan (A + B)

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Solution

The correct option is D $tan (A + B)$
$\frac{{sin}^{2} A - {sin}^{2} B}{sin A cos A - sin B cos B} = \frac{\frac{1 - cos 2 A}{2} - \frac{1 - cos 2 B}{2}}{\frac{1}{2} (sin 2 A - sin 2 B)}$
$= \frac{1 - cos 2 A - 1 + cos 2 B}{sin 2 A - sin 2 B}$
$= \frac{- (cos 2 A - cos 2 B)}{sin 2 B - sin 2 A}$
$= \frac{- {- 2 sin (\frac{2 A + 2 B}{2}) sin (\frac{2 A - 2 B}{2})}}{sin 2 B - sin 2 A}$
$== \frac{- {- 2 sin (\frac{2 A + 2 B}{2}) sin (\frac{2 A - 2 B}{2})}}{2 sin (\frac{2 A - 2 B}{2}) cos (\frac{2 A + 2 B}{2})}$
$= \frac{sin (A + B)}{cos (A + B)}$
$= tan (A + B)$

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Similar questions

\frac{{sin}^{2} A - {sin}^{2} B}{sin A cos A - sin B cos B} = tan (A - B)

Q. Prove that

\frac{{sin}^{2} A - {sin}^{2} B}{sin A cos A - sin B cos B} = tan (A + B)

\frac{cos 2 B - cos 2 A}{sin 2 A + sin 2 B} = tan (A - B)

Q. Prove that:

\frac{{sin}^{2} A - {sin}^{2} B}{sin A cos A - sin B cos B} = tan (A + B)

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Find sin2A−sin2BsinAcosA−sinBcosB=?.

Find $\frac{{sin}^{2} A - {sin}^{2} B}{sin A cos A - sin B cos B} = ?$ .