Question

If $\cos \theta =\dfrac{3}{5}$, then find the value of $\dfrac{\sin \theta -\dfrac{1}{\tan \theta }}{2\tan \theta }$.

Answer 1

Hint:First of all consider a right angled triangle ABC with C as angle $\theta$.Now as $\cos \theta =\dfrac{3}{5}$, consider base and hypotenuse as 3x and 5x respectively. Now use Pythagoras theorem to find the perpendicular side. Now find $\sin \theta =\dfrac{P}{H}$ and $\tan \theta =\dfrac{P}{B}$ and substitute in the given expression to get the required answer.

Complete step-by-step answer:
Here, we are given $\cos \theta =\dfrac{3}{5}$. We have to find the value of $\dfrac{\sin \theta -\dfrac{1}{\tan \theta }}{2\tan \theta }$.
Let us consider the expression given in the question.
$E=\dfrac{\sin \theta -\dfrac{1}{\tan \theta }}{2\tan \theta }......(1)$
Now we are given that $\cos \theta =\dfrac{3}{5}......(2)$
We know that $\cos \theta =\dfrac{base}{hypotenuse}.....(3)$
From equation (2) and (3) we get as follows:
$\dfrac{3}{5}=\dfrac{base}{hypotenuse}$
Let us assume a $\Delta ABC$, right angled at C.

Let base BC be equal to 3x and hypotenuse AC be equal to 5x.
We know that Pythagoras theorem states that in a right angled triangle, the square of the hypotenuse side is equal to the sum of the other two sides.
So in the above $\Delta ABC$ by applying the Pythagoras theorem, we get as follows:
${{\left( AB \right)}^{2}}+{{\left( BC \right)}^{2}}={{\left( AC \right)}^{2}}$
By substituting the value of BC as 3x and AC as 5x, we get as follows:

& {{\left( AB \right)}^{2}}+{{\left( 3x \right)}^{2}}={{\left( 5x \right)}^{2}} \\\ & {{\left( AB \right)}^{2}}+9{{x}^{2}}=25{{x}^{2}} \\\ & {{\left( AB \right)}^{2}}=25{{x}^{2}}-9{{x}^{2}} \\\ & {{\left( AB \right)}^{2}}=16{{x}^{2}} \\\ & AB=\sqrt{16{{x}^{2}}} \\\ & AB=4x \\\ \end{aligned}$$ So we get $$AB=4x$$. We know that $$\sin \theta =\dfrac{perpendicular}{hypotenuse}.....(4)$$ In $$\Delta ABC$$ with respect to angle $$\theta $$, Perpendicular = AB = 4x Hypotenuse = AC = 5x By substituting these values in equation (4), we get as follows: $$\sin \theta =\dfrac{4x}{5x}=\dfrac{4}{5}$$ We also know that $$\tan \theta =\dfrac{perpendicular}{base}......(5)$$ In $$\Delta ABC$$ with respect to angle $$\theta $$, Perpendicular = AB = 4x Base = BC = 3x By substituting the values in equation (5), we get as follows: $$\tan \theta =\dfrac{4x}{3x}=\dfrac{4}{3}$$ Now by substituting the values of $$\sin \theta =\dfrac{4}{5}$$ and $$\tan \theta =\dfrac{4}{3}$$ in equation (1), we get as follows: $$E=\dfrac{\sin \theta -\dfrac{1}{\tan \theta }}{2\tan \theta }$$ $$E=\dfrac{\dfrac{4}{5}-\dfrac{1}{\dfrac{4}{3}}}{2.\dfrac{4}{3}}$$ $$E=\dfrac{\dfrac{4}{5}-\dfrac{3}{4}}{\dfrac{8}{3}}$$ $$E=\dfrac{\dfrac{16-15}{20}}{\dfrac{8}{3}}$$ $$\begin{aligned} & E=\dfrac{1}{20}.\dfrac{3}{8} \\\ & E=\dfrac{3}{160} \\\ \end{aligned}$$ So we have got the value of $$\dfrac{\sin \theta -\dfrac{1}{\tan \theta }}{2\tan \theta }$$ as $$\dfrac{3}{160}$$. Note: In this type of questions, many students make mistakes while taking perpendicular and base with respect to certain angles. So they must keep in mind that the side opposite to the angle would be perpendicular and the side adjacent to the angle would be the base side. Also, the perpendicular and base get changed with respect to the angle unlike hypotenuse which is constant for a given right angled triangle.