Question

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

Answer 1

Hint: In the given expression, we don’t have any $\sec \theta$ term. For finding the value of the given expression, we have to find the values of $\sin \theta$ , $\cos \theta$ , $\tan \theta$ , and $\cot \theta$ . Value of $\cos \theta$ can be calculated by using the value of $\sec \theta$. Putting the value of $\cos \theta$ in the identity, ${{\sin }^{2}}\theta +{{\cos }^{2}}=1$, $\sin \theta$ can be calculated. Then, using the value of $\cos \theta$ and $\sin \theta$ , $\tan \theta$ and $\cot \theta$ can be calculated. Now, put the values of $\sin \theta$ , $\cos \theta$ , $\tan \theta$ , and $\cot \theta$ in the given expression and solve it further.

Complete step-by-step answer:

According to the question, it is given that $\sec \theta =\dfrac{5}{4}$………………(1)

We know the relation between $\sec \theta$ and $\cos \theta$ .

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

$\Rightarrow \cos \theta =\dfrac{1}{\sec \theta }$……………(2)

Putting the value of $\sec \theta$ in equation (2), we get

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

$\Rightarrow \cos \theta =\dfrac{4}{5}$

We also have the identity, ${{\sin }^{2}}\theta +{{\cos }^{2}}=1$ .

Taking ${{\cos }^{2}}\theta$ to RHS, we get ${{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta$………………(3)

Putting the value of $\cos \theta$ in equation (3), we get

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

$\Rightarrow {{\sin }^{2}}\theta =1-\dfrac{16}{25}$

$\Rightarrow {{\sin }^{2}}\theta =\dfrac{25-16}{25}$

$\Rightarrow {{\sin }^{2}}\theta =\dfrac{9}{25}$…………….(4)

Taking root in both LHS and RHS of the equation (4), we get

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

We also know that, $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$………………(6)

Using the values of $\sin \theta$ and $\cos \theta$ in equation (6), we get

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

$\Rightarrow \tan \theta =\dfrac{3}{4}$ ………….(7)

We also know the relation between $\tan \theta$ and $\cot \theta$,

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

$\Rightarrow \cot \theta =\dfrac{1}{\dfrac{3}{4}}$

$\Rightarrow \cot \theta =\dfrac{4}{3}$………..(8)

Now, we have the values of $\sin \theta$ , $\cos \theta$ , $\tan \theta$ , and $\cot \theta$ .

Putting these values in the expression, $\dfrac{\sin \theta -2\cos \theta }{\tan \theta -\cot \theta }$ we get,

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

$=\dfrac{\dfrac{3}{5}-2.\dfrac{4}{5}}{\dfrac{3}{4}-\dfrac{4}{3}}$

$=\dfrac{\dfrac{3-8}{5}}{\dfrac{9-16}{12}}$

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

$=\dfrac{-1}{\dfrac{-7}{12}}$

$=\dfrac{12}{7}$

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

Note: This question can also be solved using Pythagoras theorem.

$\sec \theta =\dfrac{5}{4}$

Using Pythagoras theorem, we can find height.

Height= $\sqrt{{{\left( hypotenuse \right)}^{2}}-{{\left( base \right)}^{2}}}$

$=\sqrt{{{\left( 5 \right)}^{2}}-{{\left( 4 \right)}^{2}}}$

$=\sqrt{25-16}$

$=\sqrt{9}$

$=3$

$\sin \theta =\dfrac{height}{hypotenuse}$

$\sin \theta =\dfrac{3}{5}$

$\cos \theta =\dfrac{base}{hypotenuse}$

$\cos \theta =\dfrac{4}{5}$

$\tan \theta =\dfrac{height}{base}$

$\tan \theta =\dfrac{3}{4}$

$\cot \theta =\dfrac{base}{height}$

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

Now, we have the values of $\sin \theta$ , $\cos \theta$ , $\tan \theta$ , and $\cot \theta$ .Put these values in the expression, $\dfrac{\sin \theta -2\cos \theta }{\tan \theta -\cot \theta }$ and solve it further.