Question

If \[\csc \theta +\cot \theta =P\], then prove that, \[\cos \theta =\dfrac{{{P}^{2}}-1}{{{P}^{2}}+1}\].

Answer 1

Hint: Find the inverse of \[\csc \theta \] and \[\cot \theta \]. Substitute them in the equation given and simplify it. Square the expression formed and simplify it using trigonometric identities to prove the value of \[\cos \theta \] required.

Complete step-by-step answer:
Given that, \[\csc \theta +\cot \theta =P-(1)\]
We know the basic trigonometric formulae where,
\[\csc =\dfrac{1}{\sin \theta }\]and \[\cot \theta =\dfrac{1}{\tan \theta }=\dfrac{1}{\dfrac{\sin \theta }{\cos \theta }}=\dfrac{\cos \theta }{\sin \theta }\].
Now let us substitute the value of \[\csc \theta \]and \[\cot \theta \]in equation (1).
\[\begin{align}
  & \csc \theta +\cot \theta =P \\
 & \Rightarrow \dfrac{1}{\sin \theta }+\dfrac{\cos \theta }{\sin \theta }=P \\
 & \dfrac{1+\cos \theta }{\sin \theta }=P \\
\end{align}\]
Thus we can write that, \[1+\cos \theta =P\sin \theta \].
Now let us square both sides of the expression, \[{{\left( 1+\cos \theta \right)}^{2}}={{\left( P\sin \theta \right)}^{2}}\].
We know that, \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\].
\[\therefore {{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta \]
\[\left( 1-{{\cos }^{2}}\theta \right)\]is of the form \[\left( {{a}^{2}}-{{b}^{2}} \right)=\left( a-b \right)\left( a+b \right)\].
So, \[\left( 1-{{\cos }^{2}}\theta \right)\]can be written as, \[\left( 1-\cos \theta \right)\left( 1+\cos \theta \right)\].
\[\begin{align}
  & \therefore {{\left( 1+\cos \theta \right)}^{2}}={{P}^{2}}{{\sin }^{2}}\theta \\
 & {{\left( 1+\cos \theta \right)}^{2}}={{P}^{2}}\left( 1-{{\cos }^{2}}\theta \right) \\
 & \therefore {{\left( 1+\cos \theta \right)}^{2}}={{P}^{2}}\left( 1-\cos \theta \right)\left( 1+\cos \theta \right) \\
\end{align}\]
We can cancel out \[\left( 1+\cos \theta \right)\]from LHS and RHS of the equation.
Thus we get,
\[\begin{align}
  & 1+\cos \theta ={{P}^{2}}\left( 1-\cos \theta \right) \\
 & \Rightarrow 1+\cos \theta ={{P}^{2}}-{{P}^{2}}\cos \theta \\
\end{align}\]
Let us rearrange the above expression,
\[\begin{align}
  & \cos \theta +{{P}^{2}}\cos \theta ={{P}^{2}}-1 \\
 & \cos \theta \left( 1+{{P}^{2}} \right)={{P}^{2}}-1 \\
 & \cos \theta \left( {{P}^{2}}+1 \right)={{P}^{2}}-1 \\
 & \therefore \cos \theta =\dfrac{{{P}^{2}}-1}{{{P}^{2}}+1} \\
\end{align}\]
Hence, we proved that is \[\csc \theta +\cot \theta =P\]then, \[\cos \theta =\dfrac{{{P}^{2}}-1}{{{P}^{2}}+1}\].
Hence proved.

Note: We got the expression, \[{{\left( 1+\cos \theta \right)}^{2}}={{\left( P\sin \theta \right)}^{2}}\].
Don’t use, \[{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}\]and simplify \[{{\left( 1+\cos \theta \right)}^{2}}\]. This may complicate the answer obtained. So use the trigonometric identity of \[{{\sin }^{2}}\theta \]and simplify the expression obtained.