Question

How do you verify $\dfrac{{1 + \tan x}}{{\sin x + \cos x}} = \sec x$?

Answer 1

Hint: This question is related to the trigonometry, and we have to prove that the left hand side is equal to the right hand side of the expression, and this question can be solved by using trigonometric identities i.e.,$\tan x = \dfrac{{\sin x}}{{\cos x}}$, and by further simplifying by using another trigonometric identity i.e.,$\sec x = \dfrac{1}{{\cos x}}$, we will get the result which is on the right hand of the equation.

Complete step by step answer:
Given function is $\dfrac{{1 + \tan x}}{{\sin x + \cos x}} = \sec x$,
Now we have to prove that left hand side is equal to the right hand side of the equation, now take the expression on the left hand side i.e.,
$ \Rightarrow \dfrac{{1 + \tan x}}{{\sin x + \cos x}}$,
Now using the trigonometric identity i.e., $\tan x = \dfrac{{\sin x}}{{\cos x}}$ , by writing tan in terms of sin and cos, we get,
$ \Rightarrow \dfrac{{1 + \tan x}}{{\sin x + \cos x}} = \dfrac{{1 + \dfrac{{\sin x}}{{\cos x}}}}{{\sin x + \cos x}}$,
Now simplifying by taking L.C.M we get,
$ \Rightarrow \dfrac{{1 + \tan x}}{{\sin x + \cos x}} = \dfrac{{\dfrac{{\cos x + \sin x}}{{\cos x}}}}{{\sin x + \cos x}}$,
Now taking denominator of the numerator to the denominator we get,
$ \Rightarrow \dfrac{{1 + \tan x}}{{\sin x + \cos x}} = \dfrac{{\cos x + \sin x}}{{\cos x\left( {\sin x + \cos x} \right)}}$,
Now eliminating the like terms we get,
$ \Rightarrow \dfrac{{1 + \tan x}}{{\sin x + \cos x}} = \dfrac{1}{{\cos x}}$,
Now again using the trigonometric identity, $\sec x = \dfrac{1}{{\cos x}}$, writing cos in terms of sec we get,
$ \Rightarrow \dfrac{{1 + \tan x}}{{\sin x + \cos x}} = \sec x$,
Which is equal to the right hand side of the equation,
Hence proved.

$\therefore $By using trigonometric identities we proved that expression $\dfrac{{1 + \tan x}}{{\sin x + \cos x}}$ will be equal to $\sec x$.

Note: An identity is an equation that always holds true. A trigonometric identity is an identity that contains trigonometric functions and holds true for all right-angled triangles. They are useful when solving questions with trigonometric functions and expressions. There are many trigonometric identities, here are some useful identities:
$\tan x = \dfrac{{\sin x}}{{\cos x}}$,
$\sec x = \dfrac{1}{{\cos x}}$,
$\cot x = \dfrac{1}{{\tan x}} = \dfrac{{\cos x}}{{\sin x}}$ ,
${\sin ^2}x = 1 - {\cos ^2}x$,
${\cos ^2}x + {\sin ^2}x = 1$,
${\sec ^2}x - {\tan ^2}x = 1$,
${\csc ^2}x = 1 + {\cot ^2}x$.
${\cos ^2}x - {\sin ^2}x = 1 - 2{\sin ^2}x$,
${\cos ^2}x - {\sin ^2}x = 2{\cos ^2}x - 1$,
$\sin 2x = 2\sin x\cos x$,
$2{\cos ^2}x = 1 + \cos 2x$,
$\tan 2x = \dfrac{{2\tan x}}{{1 - {{\tan }^2}x}}$.