Question

How do you simplify $\sqrt {1 + {{\tan }^2}x} $ ?

Answer 1

Hint: In the given question, we have been asked to simplify a trigonometric expression. Firstly, we will make use of the relation between squares of sine and cosine which is given by, ${\sin ^2}x + {\cos ^2}x = 1$. We also make use of the expression of tangent in terms of sine and cosine. And also we use the idea of reciprocal of cosine is nothing but secant. After using all these expressions, we simplify it and obtain the desired result.

Complete step by step answer:
Given an trigonometric expression of the form $\sqrt {1 + {{\tan }^2}x} $ …… (1)
We are asked to simplify the above expression given by the equation (1).
Firstly, we begin with the relation between the squares of sine and cosine, which is given by,
${\sin ^2}x + {\cos ^2}x = 1$
This is one of the basic formulas used in trigonometric.
Now dividing the both sides of the above equation by ${\cos ^2}x$, we get,
$ \Rightarrow \dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}} + \dfrac{{{{\cos }^2}x}}{{{{\cos }^2}x}} = \dfrac{1}{{{{\cos }^2}x}}$ …… (2)
We know the expression for tangent in terms of sine and cosine, which is given by,
$\dfrac{{\sin x}}{{\cos x}} = \tan x$
Squaring this on both sides we get,
$ \Rightarrow {\left( {\dfrac{{\sin x}}{{\cos x}}} \right)^2} = {(\tan x)^2}$
Simplifying we get,
$ \Rightarrow \dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}} = {\tan ^2}x$
Also we use the fact that reciprocal of cosine is nothing but the function secant.
i.e. $\dfrac{1}{{\cos x}} = \sec x$
Squaring on both sides we get,
${\left( {\dfrac{1}{{\cos x}}} \right)^2} = {(\sec x)^2}$
Simplifying we get,
$ \Rightarrow \dfrac{1}{{{{\cos }^2}x}} = {\sec ^2}x$
Hence the equation (2) becomes,
$ \Rightarrow {\tan ^2}x + \dfrac{{{{\cos }^2}x}}{{{{\cos }^2}x}} = {\sec ^2}x$
We know that $\dfrac{{{{\cos }^2}x}}{{{{\cos }^2}x}} = 1$
Hence we get,
$ \Rightarrow {\tan ^2}x + 1 = {\sec ^2}x$
This can also be written as,
$ \Rightarrow 1 + {\tan ^2}x = {\sec ^2}x$
Taking square root on both sides, we get,
$ \Rightarrow \sqrt {1 + {{\tan }^2}x} = \sqrt {{{\sec }^2}x} $
Simplifying this we get,
$ \Rightarrow \sqrt {1 + {{\tan }^2}x} = |\sec x|$

Hence the simplification of the expression $\sqrt {1 + {{\tan }^2}x} $ is given by $|\sec x|$.
i.e. $\sqrt {1 + {{\tan }^2}x} = |\sec x|$.

Note: Students must remember the basic formulas related to trigonometric ratios. So that they can solve such problems without much difficulty.
Some of the formulas are listed below.
(1) ${\sin ^2}x + {\cos ^2}x = 1$
(2) $1 + {\tan ^2}x = {\sec ^2}x$
(3) $1 + {\cot ^2}x = {\csc ^2}x$
(4) $\cos 2x = {\cos ^2}x - {\sin ^2}x$
(5) $\cos 2x = 2{\cos ^2}x - 1$
(6) $\cos 2x = 1 - 2{\sin ^2}x$
(7) $\cos 2x = \dfrac{{1 - {{\tan }^2}x}}{{1 + {{\tan }^2}x}}$
(8) $\sin 2x = 2\sin x\cos x$