If sin θ=35, evaluate cos θ−1tan θ2 cot θ.

Given: $\sin \theta=\frac{3}{5}$.....(1)

To find the value of $\frac{\cos \theta-\frac{1}{\tan \theta}}{2 \cot \theta}$

Now, we know the following trigonometric identity

$\cos ^{2} \theta+\sin ^{2} \theta=1$

Therefore, by substituting the value of $\sin \theta$ from equation (1),

We get,

$\cos ^{2} \theta+\left(\frac{3}{5}\right)^{2}=1$

Therefore,

$\cos ^{2} \theta=1-\left(\frac{3}{5}\right)^{2}$

$=1-\frac{(3)^{2}}{(5)^{2}}$

$=1-\frac{9}{25}$

Now by taking L.C.M

We get,

$\cos ^{2} \theta=\frac{25-9}{25}$

$=\frac{16}{25}$

Therefore by taking square root on both sides

We get,

$\cos \theta=\sqrt{\frac{16}{25}}$

$=\frac{\sqrt{16}}{\sqrt{25}}$

$=\frac{4}{5}$

Therefore,

$\cos \theta=\frac{4}{5}$.....(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 (1) and (2) respectively

We get,

$\tan \theta=\frac{\frac{3}{5}}{\frac{4}{5}}$

$\tan \theta=\frac{3}{5} \times \frac{5}{4}$

$=\frac{3}{4}$

$\tan \theta=\frac{3}{4}$...(3)

Also, we know that

$\cot \theta=\frac{1}{\tan \theta}$

Therefore from equation (4) ,

We get,

$\cot \theta=\frac{1}{\frac{3}{4}}$

$=\frac{4}{3}$

Therefore,

$\cot \theta=\frac{4}{3}$

Now, by substituting the value of $\cos \theta, \tan \theta$ and $\cot \theta$ from equation (2), (3) and (4) respectively in the expression below

$\frac{\cos \theta-\frac{1}{\tan \theta}}{2 \cot \theta}$

We get,

$\frac{\cos \theta-\frac{1}{\tan \theta}}{2 \cot \theta}=\frac{\frac{4}{5}-\frac{1}{3}}{2 \times \frac{4}{3}}$

$=\frac{\frac{4}{5}-\frac{4}{3}}{\frac{2 \times 4}{3}}$

$=\frac{\frac{4 \times 3}{5 \times 3}-\frac{4 \times 5}{3 \times 5}}{\frac{8}{3}}$

$\frac{\cos \theta-\frac{1}{\tan \theta}}{2 \cot \theta}=\frac{\frac{12}{15}-\frac{20}{15}}{\frac{8}{3}}$

$=\frac{\frac{-8}{15}}{\frac{8}{3}}$

$=\frac{-8}{15} \times \frac{3}{8}$

$=\frac{-1}{5}$

Therefore, $\frac{\cos \theta-\frac{1}{\tan \theta}}{2 \cot \theta}=\frac{-1}{5}$