If cos θ=35, find the value of sin θ−1tan θ2 tan θ.

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, $\frac{\sin \theta-\frac{1}{\tan \theta}}{2 \tan \theta}=\frac{3}{160}$