The value of left(sin 70^{circ}right)left(cot | vedclass

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Question

(A) Let $E = \sin 70^{\circ} (\cot 10^{\circ} \cot 70^{\circ} - 1)$.Using $\cot 	heta = \frac{\cos 	heta}{\sin 	heta}$,we have:$E = \sin 70^{\circ}

Vedclass · Accepted Answer

$1$

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(A) Let $E = \sin 70^{\circ} (\cot 10^{\circ} \cot 70^{\circ} - 1)$.Using $\cot 	heta = \frac{\cos 	heta}{\sin 	heta}$,we have:$E = \sin 70^{\circ} \left( \frac{\cos 10^{\circ} \cos 70^{\circ}}{\sin 10^{\circ} \sin 70^{\circ}} - 1 ight)$$E = \sin 70^{\circ} \left( \frac{\cos 10^{\circ} \cos 70^{\circ} - \sin 10^{\circ} \sin 70^{\circ}}{\sin 10^{\circ} \sin 70^{\circ}} ight)$Using the identity $\cos(A+B) = \cos A \cos B - \sin A \sin B$:$E = \sin 70^{\circ} \left( \frac{\cos(10^{\circ} + 70^{\circ})}{\sin 10^{\circ} \sin 70^{\circ}} ight)$$E = \sin 70^{\circ} \left( \frac{\cos 80^{\circ}}{\sin 10^{\circ} \sin 70^{\circ}} ight)$Since $\cos 80^{\circ} = \sin(90^{\circ} - 80^{\circ}) = \sin 10^{\circ}$:$E = \sin 70^{\circ} \left( \frac{\sin 10^{\circ}}{\sin 10^{\circ} \sin 70^{\circ}} ight) = 1$.

The value of $\left(\sin 70^{\circ}\right)\left(\cot 10^{\circ} \cot 70^{\circ}-1\right)$ is

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