Question

If $\sec \theta = \dfrac{5}{4}$, show that $\left( {\dfrac{{\sin \theta - 2\cos \theta }}{{\tan \theta - \cos \theta }}} \right) = \dfrac{{12}}{7}$?

Answer 1

In order to solve this question we have given a trigonometry relation and from that function, we are going to use the relation of sides of a right-angle triangle and then find perpendicular by using Pythagoras theorem. Then find all the required relations and put them on the left-hand side of the proof equation and solve that to get the final answer.

Complete step by step answer:
We have given the value of a trigonometric function $\sec \theta = \dfrac{5}{4}$. We have to prove $\left( {\dfrac{{\sin \theta - 2\cos \theta }}{{\tan \theta - \cos \theta }}} \right) = \dfrac{{12}}{7}$.
On the left hand side of the equation.
$\Rightarrow \left( {\dfrac{{\sin \theta - 2\cos \theta }}{{\tan \theta - \cos \theta }}} \right)$

To solve this side we must know all the values of the trigonometric function. So, now we are going to find all the values.
$\sec \theta = \dfrac{5}{4}$..........(given)
On writing the sec trigonometric function in terms of sides of a right-angle triangle.
Right angle triangle is as shown in the figure.

$\dfrac{H}{B} = \dfrac{{12}}{7}$
From this relation we use Pythagoras theorem and find the value of perpendicular.
${P^2} + {B^2} = {H^2}$
On rearranging this theorem.
$P = \sqrt {{H^2} - {B^2}}$
On putting the value of base and hypotenuse.
$P = \sqrt {25 - 16}$
On further calculating
$P = 3$
Now we are going to find all the required values that are used in the left side of the proof part.
$\sin \theta = \dfrac{3}{5}$
$\Rightarrow \cos \theta = \dfrac{4}{5}$
$\Rightarrow \tan \theta = \dfrac{3}{4}$
$\Rightarrow \left( {\dfrac{{\sin \theta - 2\cos \theta }}{{\tan \theta - \cos \theta }}} \right)$

On putting all these values in the left side of the proof part.
$\Rightarrow \left( {\dfrac{{\dfrac{3}{5} - 2\dfrac{4}{5}}}{{\dfrac{3}{4} - \dfrac{4}{5}}}} \right)$
On calculating some of this equation.
$\Rightarrow \left( {\dfrac{{\dfrac{{3 - 8}}{5}}}{{\dfrac{{15 - 16}}{{20}}}}} \right)$
On further simplifying this equation.
$\Rightarrow \left( {\dfrac{{\dfrac{{3 - 8}}{5}}}{{\dfrac{{15 - 16}}{{20}}}}} \right)$
On further solving
$\Rightarrow \left( {\dfrac{{ - 5}}{5} \times \dfrac{{20}}{{ - 1}}} \right)$
Negative is canceled by negative and 5 is canceled because this is a common factor.
$\Rightarrow 20$
Hence, the left-hand side is not equal to the right-hand side.

So the proved relation is wrong.

Note: In order to solve this question, students must have a knowledge of all the trigonometric relations in terms of sides of the right-angle triangle and theorems related to the triangle. There is another way to solve this question that is changing all the trigonometric functions in only one but there is a higher probability of committing mistakes.