If ${\tan ^2}\theta = 2{\tan ^2}\varphi + 1$ , then the value of $\cos 2\theta + {\sin ^2}\varphi $ is
A) 1
B) 2
C) 0
D) Independent of $\varphi $

Hint: In such trigonometric questions, where one equation is given and the value of another is to be found, we make changes and rearrangements in the given equation such that by using various trigonometric formulas we can reach the equation whose value is to be calculated.
Various trigonometric formulas which can be used here are:
\[
  1 + {\tan ^2}\theta = {\sec ^2}\theta \\
  \sec \theta = \dfrac{1}{{\cos \theta }} \\
  {\cos ^2}\theta + {\sin ^2}\theta = 1 \\
  \cos 2\theta = 2{\sin ^2}\theta - 1 \\
 \]
For using these, we will make changes on both sides of the equation (L.H.S and R.H.S), as changes made on only one side will change the complete meaning of the equation which is unacceptable.

Complete step-by-step answer:
Given equation is: ${\tan ^2}\theta = 2{\tan ^2}\varphi + 1$
Adding 1 on both the sides, we get:
$
  1 + {\tan ^2}\theta = 1 + 2{\tan ^2}\varphi + 1 \\
  1 + {\tan ^2}\theta = 2 + 2{\tan ^2}\varphi \\
  1 + {\tan ^2}\theta = 2\left( {1 + {{\tan }^2}\theta } \right) \\
  {\sec ^2}\theta = 2\left( {{{\sec }^2}\varphi } \right) \\
  \dfrac{1}{{{{\cos }^2}\theta }} = \dfrac{2}{{{{\cos }^2}\varphi }} \\
  {\cos ^2}\varphi = 2{\cos ^2}\theta \\
    \\
    \\
 $
\[\left[
  \because 1 + {\tan ^2}\theta = {\sec ^2}\theta \\
  \sec \theta = \dfrac{1}{{\cos \theta }} \\
\right]\]

Now subtracting 1 from both sides:
$
   - \left( {1 - {{\cos }^2}\varphi } \right) = 2{\cos ^2}\theta - 1 \\
   - {\sin ^2}\varphi = \cos 2\theta \\
  \cos 2\theta + {\sin ^2}\varphi = 0 \\
 $
 $\left[
  \because \cos 2\theta = {\sin ^2}\theta - 1 \\
  1 - {\cos ^2}\theta = {\sin ^2}\theta \\
\right]$

Therefore, it can be seen that if ${\tan ^2}\theta = 2{\tan ^2}\varphi + 1$ , then value of $\cos 2\theta + \ {\sin ^2}\varphi $ is 0, hence correct option is c).
So, the correct answer is “Option C”.

Note: There is no particular way to solve such questions, we just need to use trigonometric formulas according to the need to get the required value.
Here, we could have converted $\tan \theta $ in terms of $\sin \theta $and $\cos \theta $ in the beginning using the relation
$\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$.
Easiest way to attempt trigonometric problems is to have the knowledge about the formulas and identities, how and where these can be applied. The complex formulas need not to be crammed but can be derived from the basic ones.
Also, if we have a requirement of ${\sin ^2}\theta $ and ${\cos ^2}\theta $ we can just use them in place of 1 (generally as product) because \[{\cos ^2}\theta + {\sin ^2}\theta = 1\].