Question

If \[{{\csc }^{2}}\theta .\left( 1+\cos \theta \right)\left( 1-\cos \theta \right)=\lambda \] then the value of \[\lambda \] is
(a) 0
(b) \[{{\cos }^{2}}\theta \]
(c) 1
(d) -1

Answer 1

Hint: We solve this problem by using the standard formula of algebra and the identities of trigonometric ratios.
We have the formula of difference of squares of two numbers as
\[{{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)\]
We have the standard identity of trigonometric ratios as
\[{{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1\]

Complete step by step answer:
We are given that the equation as
\[\Rightarrow {{\csc }^{2}}\theta .\left( 1+\cos \theta \right)\left( 1-\cos \theta \right)=\lambda \]
We know that the formula of difference of squares of two numbers as
\[{{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)\]
By using the above formula in the given equation then we get
\[\begin{align}
  & \Rightarrow {{\csc }^{2}}\theta .\left( {{1}^{2}}-{{\left( \cos \theta \right)}^{2}} \right)=\lambda \\
 & \Rightarrow {{\csc }^{2}}\theta \left( 1-{{\cos }^{2}}\theta \right)=\lambda \\
\end{align}\]
We know that the standard identity of the trigonometric ratios that is
\[{{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1\]
By using this standard identity in the above equation then we get
\[\begin{align}
  & \Rightarrow {{\csc }^{2}}\theta \left( \left( {{\cos }^{2}}\theta +{{\sin }^{2}}\theta \right)-{{\cos }^{2}}\theta \right)=\lambda \\
 & \Rightarrow {{\csc }^{2}}\theta \left( {{\sin }^{2}}\theta \right)=\lambda \\
\end{align}\]
We know that the relation between the sine ratio and cosecant ratio is given as
\[\sin \theta =\dfrac{1}{\csc \theta }\]
By using this relation in above equation then we get
\[\begin{align}
  & \Rightarrow {{\csc }^{2}}\theta \left( \dfrac{1}{{{\csc }^{2}}\theta } \right)=\lambda \\
 & \Rightarrow \lambda =1 \\
\end{align}\]
Therefore we can conclude that the value of \[\lambda \] from the given equation is 1
So, option (c) is the correct answer.

Note:
We can solve this problem by using the substitution method.
We substitute some value for \[\theta \] in the given equation to get the required value.
We are given that the equation as
\[\Rightarrow {{\csc }^{2}}\theta .\left( 1+\cos \theta \right)\left( 1-\cos \theta \right)=\lambda \]
Let us assume that the value \[\theta =\dfrac{\pi }{2}\]
By substituting the value of \[\theta \] in the given equation then we get
\[\Rightarrow \lambda =\csc \left( \dfrac{\pi }{2} \right).\left( 1+\cos \dfrac{\pi }{2} \right)\left( 1-\cos \dfrac{\pi }{2} \right)\]
We know that the standard values of trigonometric ratios that is
\[\begin{align}
  & \cos \dfrac{\pi }{2}=0 \\
 & \csc \dfrac{\pi }{2}=1 \\
\end{align}\]

By substituting the required values in above equation then we get
\[\begin{align}
  & \Rightarrow \lambda =1\left( 1+0 \right)\left( 1-0 \right) \\
 & \Rightarrow \lambda =1 \\
\end{align}\]
Now, let us check the option (b)
By substituting the value \[\theta =\dfrac{\pi }{2}\] in option (b) then we get
\[\Rightarrow {{\cos }^{2}}\dfrac{\pi }{2}=0\]
But we can see that \[\lambda =1\]
So, we can say that option (b) is not the correct answer.
Therefore we can conclude that the value of \[\lambda \] from the given equation is 1
So, option (c) is the correct answer.