If $\cos \theta=\frac{3}{5}$, find the value of $\frac{\sin \theta-\frac{1}{\tan \theta}}{2 \tan \theta}$.
Given: $\cos \theta=\frac{3}{5}$...(1)
To find the value of $\frac{\sin \theta-\frac{1}{\tan \theta}}{2 \tan \theta}$
Now, we know the following trigonometric identity
$\cos ^{2} \theta+\sin ^{2} \theta=1$
Therefore, by substituting the value of $\cos \theta$ from equation (1),
We get,
$\left(\frac{3}{5}\right)^{2}+\sin ^{2} \theta=1$
Therefore,
$\sin ^{2} \theta=1-\left(\frac{3}{5}\right)^{2}$
$=1-\frac{(3)^{2}}{(5)^{2}}$
$=1-\frac{9}{25}$
$\sin ^{2} \theta=\frac{25-9}{25}$
$=\frac{16}{25}$
Therefore by taking square root on both sides
We get,
$\sin \theta=\sqrt{\frac{16}{25}}$
$=\frac{\sqrt{16}}{\sqrt{25}}$
$=\frac{4}{5}$
Therefore,
$\sin \theta=\frac{4}{5}$....(2)
Now, we know that
$\tan \theta=\frac{\sin \theta}{\cos \theta}$
Therefore by substituting the value of $\sin \theta$ and $\cos \theta$ from equation (2) and (1) respectively
We get,
$\tan \theta=\frac{\frac{4}{5}}{\frac{3}{5}}=\frac{4}{3} \ldots \ldots$(4)
Now, by substituting the value of $\sin \theta$ and $\tan \theta$ from equation (2) and (4) respectively in the expression below
$\frac{\sin \theta-\frac{1}{\tan \theta}}{2 \tan \theta}$
We get,
$\frac{\sin \theta-\frac{1}{\tan \theta}}{2 \tan \theta}=\frac{\frac{4}{5}-\frac{1}{\frac{4}{3}}}{2 \times \frac{4}{3}}$
$=\frac{\frac{4}{5}-\frac{3}{4}}{\frac{2 \times 4}{3}}$
$=\frac{\frac{4 \times 4}{5 \times 4}-\frac{3 \times 5}{4 \times 5}}{\frac{8}{3}}$
Therefore,
$\frac{\sin \theta-\frac{1}{\tan \theta}}{2 \tan \theta}=\frac{\frac{16}{20}-\frac{15}{20}}{\frac{8}{3}}$
$=\frac{\frac{1}{20}}{\frac{8}{3}}$
$=\frac{3}{160}$
Therefore, $\frac{\sin \theta-\frac{1}{\tan \theta}}{2 \tan \theta}=\frac{3}{160}$
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