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Help me, please, resolve following expression:

$$ \tan(\alpha+15^{\circ})\cdot\cot165^{\circ}, \;\;\; \mbox{if} \;\;\; \sin75^{\circ}\cdot\sin(\alpha+15^{\circ})=2\sin\alpha. $$

I try:

\begin{gather*} \tan(\alpha+15^{\circ})\cdot\cot165^{\circ}=\tan(\alpha+15^{\circ})\cdot(-\tan75^{\circ})=\\ =-\dfrac{\sin75^{\circ}\cdot\sin(\alpha+15^{\circ})}{\cos75^{\circ}\cdot\cos(\alpha+15^{\circ})}=-\dfrac{2\sin\alpha}{\cos75^{\circ}\cdot\cos(\alpha+15^{\circ})}. \end{gather*}

But what do with denominator? I don't know. Can someone tell me? Thanks.

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If $\sin(90^\circ-A)\sin(\alpha+A)=2\sin\alpha,$

$\cos A\sin(\alpha+A)=2\sin(\alpha+A-A)=2\{\sin(\alpha+A)\cos A-\cos(\alpha+A)\sin A\}$

$$\iff2\cos(\alpha+A)\sin A=\sin(\alpha+A)\cos A$$

$$\iff\dfrac{\sin(\alpha+A)}{\cos(\alpha+A)}=\dfrac{2\sin A}{\cos A}$$

$\implies\tan(\alpha+A)\cot(180^\circ-A)=2\tan A\cdot\{-\cot A\}=-2$

Here $A=15^\circ$

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From the given condition, $$\sin (\alpha+30^\circ)+\sin\alpha=4\sin \alpha \\\implies \sin(\alpha+15^{\circ})=3\sin \alpha$$ Let the required quantity is $$a=-\frac{\tan(\alpha+15^\circ)}{\tan 15^{\circ}}$$ Then, $$\frac{a+1}{1-a}=\frac{\tan 15^{\circ}-\tan(\alpha+15^\circ)}{\tan 15^{\circ}+\tan(\alpha+15^\circ)}\\=\frac{\sin\alpha}{\sin (\alpha+15^{\circ})}=\frac{1}{3}\\\implies a=-\frac{1}{2}$$

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