$sin (A + B) . sin (A - B) =$

{sin}^{2} A - {cos}^{2} B

No worries! We‘ve got your back. Try BYJU‘S free classes today!

{cos}^{2} A - {sin}^{2} B

No worries! We‘ve got your back. Try BYJU‘S free classes today!

{sin}^{2} A - {sin}^{2} B

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

{cos}^{2} A - {cos}^{2} B

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is D ${sin}^{2} A - {sin}^{2} B$
$\begin{matrix} sin (A + B) . sin (A - B) = [sin (A) cos (B) + cos (A) sin (B)] [sin (A) cos (B) - cos (A) sin (B)] = {sin}^{2} A {cos}^{2} A - {cos}^{2} A {sin}^{2} A = {sin}^{2} A [1 - {sin}^{2} B] - [1 - {sin}^{2} A] {sin}^{2} B = {sin}^{2} A - {sin}^{2} A {sin}^{2} B - {sin}^{2} B + {sin}^{2} A {sin}^{2} B = {sin}^{2} A - {sin}^{2} B H e n c e, t h e o p t i o n C i s t h e c o r r e c t a n s w e r . \end{matrix}$

Suggest Corrections

Similar questions

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

t a n^{2} A - t a n^{2} B = \frac{c o s^{2} B - c o s^{2} A}{c o s^{2} B c o s^{2} A} = \frac{s i n^{2} A - s i n^{2} B}{c o s^{2} A c o s^{2} B}

\frac{(t a n^{2} A - t a n^{2} B = (s i n^{2} A - s i n^{2} B)}{(C o s^{2} A \times C o s^{2} B)}

\frac{{cos}^{4} A}{{cos}^{2} B} + \frac{{sin}^{4} A}{{sin}^{2} B} = 1

\frac{{cos}^{4} B}{{cos}^{2} A} + \frac{{sin}^{4} B}{{sin}^{2} A} = 1

Join BYJU'S Learning Program