Prove that- tan A-B A-B -sin 2 A -sin 2 B cos 2 A -sin 2 B

On changing into sin and cos , we getLHS sin (A+B) sin (A B) #160; cos (A B) cos (A+B) #160; =sin ^ 2 A sin ^ 2 B #160; cos ^ 2 A ...

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Sign of Trigonometric Ratios in Different Quadrants

Prove that: ...

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Prove that: $\frac{tan (A + B)}{cot (A - B)} = \frac{{sin}^{2} A - {sin}^{2} B}{{cos}^{2} A - {sin}^{2} B}$

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

A + B = 90^{\circ}

\sqrt{\frac{tan A tan B + tan A cot B}{sin A sec B} - \frac{{sin}^{2} B}{{cos}^{2} A}} = tan A

{sin}^{2} A - {sin}^{2} B = sin (A + B) \times sin (A - B)

(i) (s i n^{2} A c o s^{2} B - c o s^{2} A s i n^{2} B) = (s i n^{2} A - s i n^{2} B)

(i i) (t a n^{2} A s e c^{2} B - s e c^{2} A t a n^{2} B) = (t a n^{2} A - t a n^{2} B)

{cot}^{2} A - {cot}^{2} B = \frac{{cos}^{2} A - {cos}^{2} B}{{sin}^{2} A {sin}^{2} B}

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