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What is $\frac{\sin \theta+1}{\cos \theta}$ equal to?
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Verified Answer
The correct answer is:
$\frac{\sin \theta-\cos \theta+1}{\sin \theta+\cos \theta-1}$
Consider $\frac{1+\sin \theta}{\cos \theta}=\frac{1+\frac{2 \tan \frac{\theta}{2}}{1+\tan ^{2} \frac{\theta}{2}}}{\frac{1-\tan ^{2} \frac{\theta}{2}}{1+\tan ^{2} \frac{\theta}{2}}}$
$\left(\because \sin 2 \theta=\frac{2 \tan \theta}{1+\tan ^{2} \theta}\right.$ and $\left.\cos 2 \theta=\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}\right)$
$=\frac{\left(1+\tan \frac{\theta}{2}\right)^{2}}{\left(1-\tan \frac{\theta}{2}\right)\left(1+\tan \frac{\theta}{2}\right)}$
$=\frac{1+\tan \frac{\theta}{2}}{1-\tan \frac{\theta}{2}}=\frac{\cos \frac{\theta}{2}+\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}-\sin \frac{\theta}{2}}$
Multiplied and divide by $2 \sin \frac{\theta}{2}$ $=\frac{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}+2 \sin ^{2} \frac{\theta}{2}}{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}-2 \sin ^{2} \frac{\theta}{2}}$
$=\frac{\sin \theta+1-\cos \theta}{\sin \theta-1+\cos \theta}$
$(\because \sin 2 \theta=2 \sin \theta \cos \theta$ and
$\left.\cos 2 \theta=1-2 \sin ^{2} \theta\right)$
$\left(\because \sin 2 \theta=\frac{2 \tan \theta}{1+\tan ^{2} \theta}\right.$ and $\left.\cos 2 \theta=\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}\right)$
$=\frac{\left(1+\tan \frac{\theta}{2}\right)^{2}}{\left(1-\tan \frac{\theta}{2}\right)\left(1+\tan \frac{\theta}{2}\right)}$
$=\frac{1+\tan \frac{\theta}{2}}{1-\tan \frac{\theta}{2}}=\frac{\cos \frac{\theta}{2}+\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}-\sin \frac{\theta}{2}}$
Multiplied and divide by $2 \sin \frac{\theta}{2}$ $=\frac{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}+2 \sin ^{2} \frac{\theta}{2}}{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}-2 \sin ^{2} \frac{\theta}{2}}$
$=\frac{\sin \theta+1-\cos \theta}{\sin \theta-1+\cos \theta}$
$(\because \sin 2 \theta=2 \sin \theta \cos \theta$ and
$\left.\cos 2 \theta=1-2 \sin ^{2} \theta\right)$
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