यदि sin 21^{circ} = frac{x}{y} है,तो sec 21^{ | vedclass

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(A) दिया गया है कि $\sin 21^{\circ} = \frac{x}{y}.$चूंकि $\cos 21^{\circ} = \sqrt{1 - \sin^{2} 21^{\circ}},$ इसलिए $\cos 21^{\circ} = \sqrt{1 - \frac{

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$\frac{x^{2}}{y \sqrt{y^{2}-x^{2}}}$

Vedclass · Answer

(A) दिया गया है कि $\sin 21^{\circ} = \frac{x}{y}.$चूंकि $\cos 21^{\circ} = \sqrt{1 - \sin^{2} 21^{\circ}},$ इसलिए $\cos 21^{\circ} = \sqrt{1 - \frac{x^{2}}{y^{2}}} = \frac{\sqrt{y^{2}-x^{2}}}{y}.$अब,$\sec 21^{\circ} = \frac{1}{\cos 21^{\circ}} = \frac{y}{\sqrt{y^{2}-x^{2}}}.$साथ ही,$\sin 69^{\circ} = \sin(90^{\circ} - 21^{\circ}) = \cos 21^{\circ} = \frac{\sqrt{y^{2}-x^{2}}}{y}.$अतः,$\sec 21^{\circ} - \sin 69^{\circ} = \frac{y}{\sqrt{y^{2}-x^{2}}} - \frac{\sqrt{y^{2}-x^{2}}}{y}.$लघुत्तम समापवर्त्य $y\sqrt{y^{2}-x^{2}}$ लेने पर,हमें प्राप्त होता है:$= \frac{y^{2} - (y^{2}-x^{2})}{y\sqrt{y^{2}-x^{2}}} = \frac{x^{2}}{y\sqrt{y^{2}-x^{2}}}.$

यदि $\sin 21^{\circ} = \frac{x}{y}$ है,तो $\sec 21^{\circ} - \sin 69^{\circ}$ का मान ज्ञात कीजिए।

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