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

LHS =sin (A+B) sin (A B) #160; 1 2 ( sin 2A sin 2B #160; ) #160; =tan (A+B) ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Trigonometric Ratios

Prove that: ...

Question

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

Open in App

Solution

Suggest Corrections

Similar questions

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

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

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

\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

Join BYJU'S Learning Program