Question

How do you prove the identity: $\sec x - \cos x = \sin x\tan x$ ?

Answer 1

Hint: $\sec x$ is the inverse of $\cos x$. Then, use the identity ${\sin ^2}x + {\cos ^2}x = 1$. The ratio of $\sin x$ and $\cos x$ is $\tan x$. Use this statement in proving the identity given in the question.

Complete Step by Step Solution:
We have to prove the identity - $\sec x - \cos x = \sin x\tan x$
So, taking the L.H.S. from the above equation, we get –
$ \Rightarrow \sec x - \cos x$
We know that, $\sec x$ is the inverse of $\cos x$ -
$\therefore \dfrac{1}{{\cos x}} - \cos x$
Multiplying $\cos x$ on numerator and denominator both with $\cos x$, we get –
$
   \Rightarrow \dfrac{1}{{\cos x}} - \cos x \times \dfrac{{\cos x}}{{\cos x}} \\
   \Rightarrow \dfrac{1}{{\cos x}} - \dfrac{{{{\cos }^2}x}}{{\cos x}} \\
 $
Taking $\cos x$ common from denominator and further solving, we get –
$ \Rightarrow \dfrac{1}{{\cos x}}\left( {1 - {{\cos }^2}x} \right) \cdots \left( 1 \right)$
We know that –
${\sin ^2}x + {\cos ^2}x = 1$
The above written identity can also be written as –
$ \Rightarrow 1 - {\cos ^2}x = {\sin ^2}x$
Using the above written identity in the equation (1), we get –
$ \Rightarrow \dfrac{{{{\sin }^2}x}}{{\cos x}}$
After separating $\sin x$ from ${\sin ^2}x$ in the numerator, we get –
$ \Rightarrow \sin x \times \dfrac{{\sin x}}{{\cos x}} \cdots \left( 2 \right)$
We know that the ratio of $\sin x$ and $\cos x$ is known as $\tan x$. Expressing this in the mathematical form, we get –
$\tan x = \dfrac{{\sin x}}{{\cos x}}$
Putting $\tan x = \dfrac{{\sin x}}{{\cos x}}$ in the equation (2), we get –
$ \Rightarrow \sin x\tan x$
$\therefore \sec x - \cos x = \sin x\tan x$
Hence, proved

Note: Many students make mistakes while putting the identities of trigonometry in the wrong places and their answer will be wrong. This can also be solved by using the triangle and Pythagoras theorem.