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If $\tan \theta+\sec \theta=4$, then what is the value of $\sin \theta ?
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The correct answer is:
$15 / 17$
$\begin{aligned} & \tan \theta+\sec \theta=4 \\ & \Rightarrow \frac{\sin \theta}{\cos \theta}+\frac{1}{\cos \theta}=4 \\ & \Rightarrow 1+\sin \theta=4 \cos \theta \end{aligned}$
Squaring on both side, $(1+\sin \theta)^{2}=16 \cos ^{2} \theta=16\left(1-\sin ^{2} \theta\right)$
$(1+\sin \theta)^{2}=16(1-\sin \theta)(1+\sin \theta)$
$1+\sin \theta=16-16 \sin \theta$
$17 \sin \theta=16-1$
$\sin \theta=\frac{15}{17}$
Squaring on both side, $(1+\sin \theta)^{2}=16 \cos ^{2} \theta=16\left(1-\sin ^{2} \theta\right)$
$(1+\sin \theta)^{2}=16(1-\sin \theta)(1+\sin \theta)$
$1+\sin \theta=16-16 \sin \theta$
$17 \sin \theta=16-1$
$\sin \theta=\frac{15}{17}$
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