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:
\[\begin{align}
  & {{\left( 3x \right)}^{2}}+{{\left( BC \right)}^{2}}={{\left( 5x \right)}^{2}} \\
 & 9{{x}^{2}}+{{\left( BC \right)}^{2}}=25{{x}^{2}} \\
 & {{\left( BC \right)}^{2}}=25{{x}^{2}}-9{{x}^{2}} \\
 & {{\left( BC \right)}^{2}}=16{{x}^{2}} \\
 & BC=\sqrt{16{{x}^{2}}} \\
 & BC=4x \\
\end{align}\]
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:
\[E=\dfrac{\cos \theta -\dfrac{1}{\tan \theta }}{2\cot \theta }\]
\[E=\dfrac{\dfrac{4}{5}-\dfrac{1}{\dfrac{3}{4}}}{2.\dfrac{4}{3}}\]
\[E=\dfrac{\dfrac{4}{5}-\dfrac{4}{3}}{2.\dfrac{4}{3}}\]
\[E=\dfrac{\dfrac{12-20}{15}}{\dfrac{8}{3}}\]
\[\begin{align}
  & E=-\dfrac{8}{15}.\dfrac{3}{8} \\
 & E=-\dfrac{1}{5} \\
\end{align}\]
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 \[\dfrac{\pi }{2}\].