frac{cos 15^{circ} cos^2 22frac{1}{2}^{circ} | vedclass

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Question

(C) माना व्यंजक $E = \frac{\cos 15^{\circ} \cos^2 22.5^{\circ} - \sin 75^{\circ} \sin^2 52.5^{\circ}}{\cos^2 15^{\circ} - \cos^2 75^{\circ}}$ है।चूंकि

Vedclass · Accepted Answer

$\frac{1}{4}$

Vedclass · Answer

(C) माना व्यंजक $E = \frac{\cos 15^{\circ} \cos^2 22.5^{\circ} - \sin 75^{\circ} \sin^2 52.5^{\circ}}{\cos^2 15^{\circ} - \cos^2 75^{\circ}}$ है।चूंकि $\sin 75^{\circ} = \cos 15^{\circ}$,अंश $\cos 15^{\circ} (\cos^2 22.5^{\circ} - \sin^2 52.5^{\circ})$ हो जाता है।सर्वसमिका $\cos^2 A - \sin^2 B = \cos(A+B) \cos(A-B)$ का उपयोग करने पर,$\cos^2 22.5^{\circ} - \sin^2 52.5^{\circ} = \cos(75^{\circ}) \cos(-30^{\circ}) = \cos 75^{\circ} \cos 30^{\circ}$ प्राप्त होता है।अतः,अंश $\cos 15^{\circ} \cos 75^{\circ} \cos 30^{\circ}$ है।चूंकि $\cos 75^{\circ} = \sin 15^{\circ}$,अंश $\cos 15^{\circ} \sin 15^{\circ} \cos 30^{\circ} = \frac{1}{2} \sin 30^{\circ} \cos 30^{\circ} = \frac{1}{4} \sin 60^{\circ} = \frac{\sqrt{3}}{8}$ है।हर $\cos^2 15^{\circ} - \cos^2 75^{\circ} = \cos^2 15^{\circ} - \sin^2 15^{\circ} = \cos 30^{\circ} = \frac{\sqrt{3}}{2}$ है।इस प्रकार,$E = \frac{\frac{\sqrt{3}}{8}}{\frac{\sqrt{3}}{2}} = \frac{1}{4}$।

$\frac{\cos 15^{\circ} \cos^2 22\frac{1}{2}^{\circ} - \sin 75^{\circ} \sin^2 52\frac{1}{2}^{\circ}}{\cos^2 15^{\circ} - \cos^2 75^{\circ}} = $

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