The value of sin ^{2} 75^{circ}-sin ^{2} 15^{ | vedclass

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Question

(A) We use the trigonometric identity $\sin^{2} A - \sin^{2} B = \sin(A+B) \sin(A-B)$.Here,$A = 75^{\circ}$ and $B = 15^{\circ}$.Substituting these va

Vedclass · Accepted Answer

$\frac{\sqrt{3}}{2}$

Vedclass · Answer

(A) We use the trigonometric identity $\sin^{2} A - \sin^{2} B = \sin(A+B) \sin(A-B)$.Here,$A = 75^{\circ}$ and $B = 15^{\circ}$.Substituting these values into the identity:$= \sin(75^{\circ} + 15^{\circ}) \sin(75^{\circ} - 15^{\circ})$$= \sin(90^{\circ}) \sin(60^{\circ})$Since $\sin(90^{\circ}) = 1$ and $\sin(60^{\circ}) = \frac{\sqrt{3}}{2}$,$= 1 	imes \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}$.

The value of $\sin ^{2} 75^{\circ}-\sin ^{2} 15^{\circ}$ is:

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