यदि 1-cot 23^{circ}=frac{x}{1-cot 22^{circ}} | vedclass

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(B) दिया गया है,$1-\cot 23^{\circ}=\frac{x}{1-\cot 22^{\circ}}$$x = (1-\cot 23^{\circ})(1-\cot 22^{\circ})$$x = (1 - \frac{\cos 23^{\circ}}{\sin 23^{\

Vedclass · Accepted Answer

$2$

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(B) दिया गया है,$1-\cot 23^{\circ}=\frac{x}{1-\cot 22^{\circ}}$$x = (1-\cot 23^{\circ})(1-\cot 22^{\circ})$$x = (1 - \frac{\cos 23^{\circ}}{\sin 23^{\circ}})(1 - \frac{\cos 22^{\circ}}{\sin 22^{\circ}})$$x = \frac{(\sin 23^{\circ} - \cos 23^{\circ})(\sin 22^{\circ} - \cos 22^{\circ})}{\sin 23^{\circ} \sin 22^{\circ}}$$x = \frac{\sin 23^{\circ} \sin 22^{\circ} - \sin 23^{\circ} \cos 22^{\circ} - \cos 23^{\circ} \sin 22^{\circ} + \cos 23^{\circ} \cos 22^{\circ}}{\sin 23^{\circ} \sin 22^{\circ}}$$\cos(A-B) = \cos A \cos B + \sin A \sin B$ और $\sin(A+B) = \sin A \cos B + \cos A \sin B$ का उपयोग करने पर:$x = \frac{\cos(23^{\circ}-22^{\circ}) - \sin(23^{\circ}+22^{\circ})}{\sin 23^{\circ} \sin 22^{\circ}}$$x = \frac{\cos 1^{\circ} - \sin 45^{\circ}}{\sin 23^{\circ} \sin 22^{\circ}}$$2 \sin A \sin B = \cos(A-B) - \cos(A+B)$ का उपयोग करने पर:$x = \frac{2(\cos 1^{\circ} - \sin 45^{\circ})}{\cos(23^{\circ}-22^{\circ}) - \cos(23^{\circ}+22^{\circ})}$$x = \frac{2(\cos 1^{\circ} - \frac{1}{\sqrt{2}})}{\cos 1^{\circ} - \cos 45^{\circ}} = \frac{2(\cos 1^{\circ} - \frac{1}{\sqrt{2}})}{\cos 1^{\circ} - \frac{1}{\sqrt{2}}} = 2$

यदि $1-\cot 23^{\circ}=\frac{x}{1-\cot 22^{\circ}}$ है,तो $x=$

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