If 1-cot 23^{circ}=frac{x}{1-cot 22^{circ}},t | vedclass

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Question

(B) Given,$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}}

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) Given,$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}}$Using $\cos(A-B) = \cos A \cos B + \sin A \sin B$ and $\sin(A+B) = \sin A \cos B + \cos A \sin B$:$x = \frac{\cos(23^{\circ}-22^{\circ}) - \sin(23^{\circ}+22^{\circ})}{\sin 23^{\circ} \sin 22^{\circ}}$$x = \frac{\cos 1^{\circ} - \sin 45^{\circ}}{\sin 23^{\circ} \sin 22^{\circ}}$Using $2 \sin A \sin B = \cos(A-B) - \cos(A+B)$:$x = \frac{2(\cos 1^{\circ} - \sin 45^{\circ})}{\cos(23^{\circ}-22^{\circ}) - \cos(23^{\circ}+22^{\circ})}$$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$

If $1-\cot 23^{\circ}=\frac{x}{1-\cot 22^{\circ}}$,then $x=$

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