Question:
If $\sin \theta=\frac{c}{\sqrt{c^{2}+d^{2}}}$, where $d>0$ then find the values of $\cos \theta$ and $\tan \theta$.
Solution:
Given : $\sin \theta=\frac{c}{\sqrt{c^{2}+d^{2}}}$
Since, $\sin \theta=\frac{P}{H}$
$\Rightarrow P=c$ and $H=\sqrt{c^{2}+d^{2}}$
Using Pythagoras theorem,
$P^{2}+B^{2}=H^{2}$
$\Rightarrow c^{2}+B^{2}=c^{2}+d^{2}$
$\Rightarrow B^{2}=d^{2}$
$\Rightarrow B=d$
Therefore,
$\cos \theta=\frac{B}{H}=\frac{d}{\sqrt{c^{2}+d^{2}}}$
$\tan \theta=\frac{P}{B}=\frac{c}{d}$
Hence, $\cos \theta=\frac{d}{\sqrt{c^{2}+d^{2}}}$ and $\tan \theta=\frac{c}{d}$.
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