$\cos ^2 5^{\circ}-\cos ^2 15^{\circ}-\sin ^2 15^{\circ}+\sin ^2 35^{\circ}$ $+\cos 15^{\circ} \sin 15^{\circ}-\cos 5^{\circ} \sin 35^{\circ}=$

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Aicharya · Accepted Answer

\(\begin{aligned} & \cos ^2 5^{\circ}-\cos ^2 15^{\circ}-\sin ^2 15^{\circ}+\sin ^2 35^{\circ} \ & +\cos 15^{\circ} \sin 15^{\circ}-\cos 5^{\circ} \sin 35^{\circ} \ & =\cos 5^{\circ}\left(\cos 5^{\circ}-\sin 35^{\circ}\right)-\cos 15^{\circ}\left(\cos 15^{\circ}-\sin 15^{\circ}\right) \ & +\sin ^2 35^{\circ}-\sin ^2 15^{\circ} \ & =\cos 5^{\circ}\left(\cos 5^{\circ}-\cos 55^{\circ}\right)-\cos 15^{\circ}\left(\cos 15^{\circ}-\cos 75^{\circ}\right) \ & +\sin 50^{\circ} \sin 20^{\circ} \ & =\cos 5^{\circ}\left(2 \sin 30^{\circ} \sin 25^{\circ}\right)-\cos 15^{\circ}\left(2 \sin 45^{\circ} \sin 30^{\circ}\right) \ & +\sin 50^{\circ} \sin 20^{\circ} \ & =-\frac{\cos 15^{\circ}}{\sqrt{2}}+\cos 5^{\circ} \sin 25^{\circ}+\sin 50^{\circ} \sin 20^{\circ} \ & =-\frac{\cos 15^{\circ}}{\sqrt{2}}+\frac{1}{2}\left(2 \cos 5^{\circ} \cos 65^{\circ}\right)+\frac{1}{2}\left(2 \sin 50^{\circ} \sin 20^{\circ}\right) \ & =-\frac{\cos 15^{\circ}}{\sqrt{2}}+\frac{1}{2}\left[\cos 70^{\circ}+\cos 60^{\circ}+\cos 30^{\circ}-\cos 70^{\circ}\right] \ & =-\frac{\cos 15^{\circ}}{\sqrt{2}}+\frac{1}{2}\left[2 \cos 45^{\circ} \cos 15^{\circ}\right] \ & =-\frac{\cos 15^{\circ}}{\sqrt{2}}+\frac{\cos 15^{\circ}}{\sqrt{2}}=0 \ \end{aligned}\) Hence, option (1) is correct.

\(\cos ^2 5^{\circ}-\cos ^2 15^{\circ}-\sin ^2 15^{\circ}+\sin ^2 35^{\circ}\) \(+\cos 15^{\circ} \sin…

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