Question

How do you prove $1 + {\tan ^2}\left( x \right) = {\sec ^2}\left( x \right)$?

Answer 1

In the given question, we have been given to prove a standard trigonometric identity. To do that, we write the basic identity of the sum of squares of sine and cosine. Then we make substitutions and get the value.

Formula Used:
We are going to use the basic identity of the sum of the square of sine and cosine, which is:
${\sin ^2}x + {\cos ^2}x = 1$

Complete step by step answer:
Let us first consider the identity of sum of square of sine and cosine,
${\sin ^2}x + {\cos ^2}x = 1$
Dividing both sides by ${\cos ^2}x$,
$\dfrac{{{{\cos }^2}x}}{{{{\cos }^2}x}} + \dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}} = \dfrac{1}{{{{\cos }^2}x}}$
We know,
$\tan x = \dfrac{{\sin x}}{{\cos x}} \Rightarrow {\tan ^2}x = \dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}}$
and also, $\sec x = \dfrac{1}{{\cos x}} \Rightarrow {\sec ^2}x = \dfrac{1}{{{{\cos }^2}x}}$
Hence, $1 + {\tan ^2}x = {\sec ^2}x$
Hence, proved.

Additional Information:
Just like the identity of tangent and secant, we have an identity of cotangent and cosecant too,
$1 + {\cot ^2}\left( x \right) = {{\mathop{\rm cosec}\nolimits} ^2}\left( x \right)$

Note: In this question, we had to prove the identity of the sum of squares of tangent and secant. We did that by writing the basic identity of a sum of the square of sine and cosine, then dividing the identity by the square of sine, and then applying the substitutions to prove the value. So, it is really important that we know the formulae and where, when, and how to use them so that we can get the correct result.