Question

If \[{{\cos }^{2}}\theta -{{\sin }^{2}}\theta ={{\tan }^{2}}\phi \]. Prove that \[\cos \phi =\dfrac{1}{\sqrt{2}\cos \theta }\].

Answer 1

Hint: In order to prove \[\cos \phi =\dfrac{1}{\sqrt{2}\cos \theta }\], firstly we will be applying the trigonometric identities. And then for easy calculation and to bring it to the form of identities, we will be adding one to both sides of the equation. And solving them accordingly will give us the required answer.

Complete step-by-step answer:
Now let us have a brief regarding the trigonometric identities. Generally, trigonometric identities are the equalities that occur for the trigonometric ratios and are true for all of the occurring variables. These trigonometric identities contain functions of one or more than one angles. There are six trigonometric ratios, they are sine, cosine, tangent, cotangent, secant and cosecant. These ratios are generally expressed as the ratios of the sides of a right angled triangle.
Now let us prove that if \[{{\cos }^{2}}\theta -{{\sin }^{2}}\theta ={{\tan }^{2}}\phi \], then \[\cos \phi =\dfrac{1}{\sqrt{2}\cos \theta }\].
We are given with \[{{\cos }^{2}}\theta -{{\sin }^{2}}\theta ={{\tan }^{2}}\phi \].
So consider \[{{\cos }^{2}}\theta -{{\sin }^{2}}\theta ={{\tan }^{2}}\phi \]
Now let us add \[1\] on both sides of the equation. Then we get,
\[{{\cos }^{2}}\theta -{{\sin }^{2}}\theta +1={{\tan }^{2}}\phi +1\]
Upon rearranging the terms in order to obtain the identity, we get
\[{{\cos }^{2}}\theta +\left( 1-{{\sin }^{2}}\theta \right)=\left( {{\tan }^{2}}\phi +1 \right)\]
We know that \[\left( 1-{{\sin }^{2}}\theta \right)\]\[=\cos \theta \] and also \[\left( {{\tan }^{2}}\phi +1 \right)\]\[={{\sec }^{2}}\phi \]
On replacing these values we get,
\[\begin{align}
  & {{\cos }^{2}}\theta +\left( 1-{{\sin }^{2}}\theta \right)=\left( {{\tan }^{2}}\phi +1 \right) \\
 & {{\cos }^{2}}\theta +{{\cos }^{2}}\theta ={{\sec }^{2}}\phi \\
\end{align}\]
On further solving,
\[\begin{align}
  & \Rightarrow 2{{\cos }^{2}}\theta ={{\sec }^{2}}\phi \\
 & \Rightarrow 2{{\cos }^{2}}\theta {{\cos }^{2}}\phi =1 \\
\end{align}\]
Since we are supposed to prove with the \[\cos \phi \] term, we will be transposing the rest of the terms except \[\cos \phi \] to the right hand side.
Doing this, we get
\[\begin{align}
  & {{\cos }^{2}}\phi =\dfrac{1}{2{{\cos }^{2}}\theta } \\
 & \cos \phi =\dfrac{1}{\sqrt{2}{{\cos }^{2}}\theta } \\
\end{align}\]
Hence proved

Note: In order to solve such trigonometric functions, we should hold a grip upon the identities and try to change the correct term so as to obtain the required solution. We can use these trigonometric identities for solving such functions as well as we can apply them for geometry too.