Question

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

Answer 1

Hint: First of all consider a right angled triangle ABC with C as angle $\theta$.Now as $\sin \theta =\dfrac{3}{5}$, consider perpendicular and hypotenuse as 3x and 5x respectively. Now use Pythagoras theorem to find the perpendicular side. Now find $cos\theta =\dfrac{B}{H}$ and $\tan \theta =\dfrac{P}{B}=\dfrac{1}{\cot \theta }$ and substitute in the given expression to get the required answer.

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

Let perpendicular AB 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 AB as 3x and AC as 5x, we get as follows:

& {{\left( 3x \right)}^{2}}+{{\left( BC \right)}^{2}}={{\left( 5x \right)}^{2}} \\\ & 9{{x}^{2}}+{{\left( BC \right)}^{2}}=25{{x}^{2}} \\\ & {{\left( BC \right)}^{2}}=25{{x}^{2}}-9{{x}^{2}} \\\ & {{\left( BC \right)}^{2}}=16{{x}^{2}} \\\ & BC=\sqrt{16{{x}^{2}}} \\\ & BC=4x \\\ \end{aligned}$$ So we get $$BC=4x$$. We know that $$\cos \theta =\dfrac{base}{hypotenuse}.....(4)$$ In $$\Delta ABC$$ with respect to angle $$\theta $$, Base = BC = 4x Hypotenuse = AC = 5x By substituting these values in equation (4), we get as follows: $$\cos \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 $$, Perpendicular = AB = 3x Base = BC = 4x By substituting the values in equation (5), we get as follows: $$\tan \theta =\dfrac{3x}{4x}=\dfrac{3}{4}$$ We also know that $$cot\theta =\dfrac{1}{\tan \theta }=\dfrac{1}{\dfrac{3}{4}}=\dfrac{4}{3}$$ Now by substituting the values of $$\cos \theta =\dfrac{4}{5}$$, $$\tan \theta =\dfrac{3}{4}$$ and $$\cot \theta =\dfrac{4}{3}$$ in equation (1), we get as follows: $$E=\dfrac{\cos \theta -\dfrac{1}{\tan \theta }}{2\cot \theta }$$ $$E=\dfrac{\dfrac{4}{5}-\dfrac{1}{\dfrac{3}{4}}}{2.\dfrac{4}{3}}$$ $$E=\dfrac{\dfrac{4}{5}-\dfrac{4}{3}}{2.\dfrac{4}{3}}$$ $$E=\dfrac{\dfrac{12-20}{15}}{\dfrac{8}{3}}$$ $$\begin{aligned} & E=-\dfrac{8}{15}.\dfrac{3}{8} \\\ & E=-\dfrac{1}{5} \\\ \end{aligned}$$ So we have got the value of $$\dfrac{\cos \theta -\dfrac{1}{\tan \theta }}{2\cot \theta }$$ as $$-\dfrac{1}{5}$$. Note: In this type of questions, students can also find the value of $$\cos \theta $$ by using the formula $$si{{n}^{2}}\theta +{{\cos }^{2}}\theta =1$$. After getting $$\cos \theta $$, students can find the value of $$\tan \theta $$ by using $$\tan \theta =\dfrac{\sin \theta }{\cos \theta }$$ and $$\cot \theta $$ by using $$\cot \theta =\dfrac{\cos \theta }{\sin \theta }$$. Also, when nothing about an angle is given in the question, assume it to be in the first quadrant that is between 0 and $$\dfrac{\pi }{2}$$.