Question

If the value of \[\cot \theta =\dfrac{1}{\sqrt{3}}\] , show that \[\dfrac{1-{{\cos }^{2}}\theta }{2-{{\sin }^{2}}\theta }=\dfrac{3}{5}\] .

Answer 1

Hint: In the question, the value of \[\cot \theta \] is given. We have to find the value of
\[\dfrac{1-{{\cos }^{2}}\theta }{2-{{\sin }^{2}}\theta }\]. In the given expression, we don’t have any \[\cot \theta \] term. For finding the value of the given expression, we have to make the expression in the form of \[\cot \theta \]. Using the identity \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\] , we can write \[1-{{\cos }^{2}}\theta ={{\sin }^{2}}\theta \]. Then divide by
\[{{\sin }^{2}}\theta \] in both numerator and denominator. Use the identity, \[{{\operatorname{cosec}}^{2}}\theta -1={{\cot }^{2}}\theta \] and solve it further.

Let us proceed with the LHS of the given expression.
In LHS we have, \[\dfrac{1-{{\cos }^{2}}\theta }{2-{{\sin }^{2}}\theta }\]……….(1)
We know that,
\[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]…………(2)
Taking \[{{\cos }^{2}}\theta \] to RHS in equation(2), we get
\[{{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta \]……….(3)
From equation (3) and equation (1), we get
\[\begin{align}
  & \dfrac{1-{{\cos }^{2}}\theta }{2-{{\sin }^{2}}\theta } \\
 & =\dfrac{{{\sin }^{2}}\theta }{2-{{\sin }^{2}}\theta } \\
\end{align}\]
Now, we have \[\dfrac{{{\sin }^{2}}\theta }{2-{{\sin }^{2}}\theta }\]…………………(4)
Dividing by \[{{\sin }^{2}}\theta \] in numerator and denominator, we get
\[\dfrac{1}{2\cos e{{c}^{2}}\theta -1}\]…………………….(5)
Equation (5) can also be written as,
\[\dfrac{1}{2(cose{{c}^{2}}\theta -1)+1}\]……………………(6)
We know the relation between \[\operatorname{cosec}\theta \] and \[\cot \theta \] .
\[{{\operatorname{cosec}}^{2}}\theta -{{\cot }^{2}}\theta =1\]………………(7)
Taking \[{{\operatorname{cosec}}^{2}}\theta \] to RHS in equation(7), we get
\[{{\cot }^{2}}\theta ={{\operatorname{cosec}}^{2}}\theta -1\]…………………..(8)
Using equation (8), we can write equation (6) as
\[\dfrac{1}{2(cose{{c}^{2}}\theta -1)+1}\]
\[=\dfrac{1}{2(co{{t}^{2}}\theta )+1}\]……………(9)
According to the question, it is given that
\[\cot \theta =\dfrac{1}{\sqrt{3}}\]……………(10)
Putting the value of \[\cot \theta \] from equation (10) in equation (9), we get
\[\begin{align}
  & =\dfrac{1}{2{{\left( \dfrac{1}{\sqrt{3}} \right)}^{2}}+1} \\
 & =\dfrac{1}{\dfrac{2}{3}+1} \\
 & =\dfrac{1}{\dfrac{2+3}{3}} \\
 & =\dfrac{3}{5} \\
\end{align}\]
So, \[\dfrac{1-{{\cos }^{2}}\theta }{2-{{\sin }^{2}}\theta }=\dfrac{3}{5}\] .
Therefore, LHS=RHS.
Hence, proved.

Note: This question can also be solved by putting the value of \[\cos \theta \] and \[\sin \theta \] . But the values of \[\cos \theta \] and \[\sin \theta \] is not given in the question. So, by using the value of \[\cot \theta \] , we can find the value of \[\cos \theta \] and \[\sin \theta \] .

\[\cot \theta =\dfrac{1}{\sqrt{3}}\]
Using Pythagoras theorem, we can find hypotenuse.
Hypotenuse= \[\sqrt{{{\left( height \right)}^{2}}+{{\left( base \right)}^{2}}}\]
\[\begin{align}
  & =\sqrt{{{\left( \sqrt{3} \right)}^{2}}+{{\left( 1 \right)}^{2}}} \\
 & =\sqrt{3+1} \\
 & =\sqrt{4} \\
 & =2 \\
\end{align}\]
\[\cos \theta =\dfrac{1}{2}\] and \[\sin \theta =\dfrac{\sqrt{3}}{2}\] .
And after putting the value of \[\sin \theta \] and \[\cos \theta \] in the given expression, we can get our required answer.