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If $\operatorname{cosec} \theta+\cot \theta=\mathrm{c}$, then what is $\cos \theta$ equal to $?
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The correct answer is:
$\frac{\mathrm{c}^{2}-1}{\mathrm{c}^{2}+1}$
Let $\operatorname{cosec} \theta+\cot \theta=\mathrm{c}$
$\Rightarrow \frac{1}{\sin \theta}+\frac{\cos \theta}{\sin \theta}=\mathrm{c} \Rightarrow \frac{1+\cos \theta}{\sin \theta}=\mathrm{c}$
$\Rightarrow \frac{1+\left(2 \cos ^{2} \frac{\theta}{2}-1\right)}{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}=\mathrm{c} \Rightarrow \frac{2 \cos ^{2} \frac{\theta}{2}}{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}=\mathrm{c}$
$\Rightarrow \cot \frac{\theta}{2}=\mathrm{c} \Rightarrow \cos \theta=\frac{1-\frac{1}{\mathrm{c}^{2}}}{1+\frac{1}{\mathrm{c}^{2}}}=\frac{\mathrm{c}^{2}-1}{\mathrm{c}^{2}+1}$
$\left(\because \cos \theta=\frac{1-\tan ^{2}\left(\frac{\theta}{2}\right)}{1+\tan ^{2}\left(\frac{\theta}{2}\right)}\right)$
$\Rightarrow \tan \frac{\theta}{2}=\frac{1}{\mathrm{c}}$
$\Rightarrow \frac{1}{\sin \theta}+\frac{\cos \theta}{\sin \theta}=\mathrm{c} \Rightarrow \frac{1+\cos \theta}{\sin \theta}=\mathrm{c}$
$\Rightarrow \frac{1+\left(2 \cos ^{2} \frac{\theta}{2}-1\right)}{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}=\mathrm{c} \Rightarrow \frac{2 \cos ^{2} \frac{\theta}{2}}{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}=\mathrm{c}$
$\Rightarrow \cot \frac{\theta}{2}=\mathrm{c} \Rightarrow \cos \theta=\frac{1-\frac{1}{\mathrm{c}^{2}}}{1+\frac{1}{\mathrm{c}^{2}}}=\frac{\mathrm{c}^{2}-1}{\mathrm{c}^{2}+1}$
$\left(\because \cos \theta=\frac{1-\tan ^{2}\left(\frac{\theta}{2}\right)}{1+\tan ^{2}\left(\frac{\theta}{2}\right)}\right)$
$\Rightarrow \tan \frac{\theta}{2}=\frac{1}{\mathrm{c}}$
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