Question

If $\sec \theta =\frac{5}{4}$, find the value of $\frac{\sin \theta -2\cos \theta }{\tan \theta -\cot \theta }$.

Answer 1

Given: ……(1)

To find the value of

Now we know that

Therefore,

Therefore from equation (1)

…… (2)

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

Therefore,

Substituting the value of from equation (2)

We get,

Therefore

…… (3)

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

Therefore,

${\tan }^2\theta =se{c}^2\theta -1$

Therefore

$$\begin{gathered} {\tan }^2\theta ={\left(\frac{5}{4}\right)}^2-1 \\ =\frac{25}{16}-1 \\ =\frac{9}{16} \end{gathered}$$

Therefore,

$$\begin{gathered} \tan \theta =\sqrt{\frac{9}{16}} \\ =\frac{3}{4} \end{gathered}$$

Therefore,

$\tan \theta =\frac{3}{4}$ …… (4)

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

Therefore, from equation (4)

We get,

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

$cot\theta =\frac{4}{3}$…… (5)

Substituting the value of,,and from equation (2) (3) (4) and (5) respectively in the expression below

We get,

$$\begin{gathered} \frac{\sin \theta -2\cos \theta }{\tan \theta -\cot \theta }=\frac{{\displaystyle \frac{3}{5}}-2\left({\displaystyle \frac{4}{5}}\right)}{{\displaystyle \frac{3}{4}}-{\displaystyle \frac{4}{3}}} \\ =\frac{{\displaystyle \frac{3}{5}}-{\displaystyle \frac{8}{5}}}{{\displaystyle \frac{\left(3\times 3\right)-\left(4\times 4\right)}{4\times 3}}} \\ =\frac{{\displaystyle \frac{3}{5}}-{\displaystyle \frac{8}{5}}}{{\displaystyle \frac{\left(3\times 3\right)-\left(4\times 4\right)}{4\times 3}}} \\ =\frac{{\displaystyle \frac{3-8}{5}}}{{\displaystyle \frac{9-16}{4\times 3}}} \\ =\frac{{\displaystyle \frac{-5}{5}}}{{\displaystyle \frac{-7}{12}}} \\ =\frac{12}{7} \end{gathered}$$

Therefore,