If sec θ=54, find the value of sin θ−2 cos θtan θ−cot θ.

Solution:

Given: $\sec \theta=\frac{5}{4}$....(1)

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

Now we know that $\sec \theta=\frac{1}{\cos \theta}$

Therefore,

$\cos \theta=\frac{1}{\sec \theta}$

Therefore from equation (1)

$\cos \theta=\frac{1}{\frac{5}{4}}$

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

Also, we know that $\cos ^{2} \theta+\sin ^{2} \theta=1$

Therefore,

$\sin ^{2} \theta=I-\cos ^{2} \theta$

$\sin \theta=\sqrt{1-\cos ^{2} \theta}$

Substituting the value of $\cos \theta$ from equation (2)

We get,

$\sin \theta=\sqrt{1-\left(\frac{4}{5}\right)^{2}}$

$=\sqrt{1-\frac{4^{2}}{5^{2}}}$

$=\sqrt{1-\frac{16}{25}}$

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

$=\sqrt{\frac{9}{25}}$

$=\frac{3}{5}$

Therefore

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

Also, we know that $\sec ^{2} \theta=1+\tan ^{2} \theta$.

Therefore,

$\tan ^{2} \theta=\sec ^{2} \theta-1$

Therefore

$\tan ^{2} \theta=\left(\frac{5}{4}\right)^{2}-1$

$=\frac{25}{16}-1$

$=\frac{9}{16}$

Therefore,

$\tan \theta=\sqrt{\frac{9}{16}}$

$=\frac{3}{4}$

Therefore,

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

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

Therefore, from equation (4)

We get,

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

$\cot \theta=\frac{4}{3}$.....(5)

Substituting the value of $\cos \theta, \sin \theta, \cot \theta$ and $\tan \theta$ from equation (2) (3) (4) and (5) respectively in the expression below

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

We get,

$\frac{\sin \theta-2 \cos \theta}{\tan \theta-\cot \theta}=\frac{\frac{3}{5}-2\left(\frac{4}{5}\right)}{\frac{3}{4}-\frac{4}{3}}$

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

$=\frac{\frac{\frac{3}{5}-\frac{3}{5}}{(3 \times 3)-(4 \times 4)}}{4 \times 3}$

$=\frac{\frac{3-8}{5}}{\frac{9-16}{4 \times 3}}$

$=\frac{\frac{-5}{5}}{\frac{-7}{12}}$

$=\frac{12}{7}$

Therefore, $\frac{\sin \theta-2 \cos \theta}{\tan \theta-\cot \theta}=\frac{12}{7}$