Question

How do you prove: $\cos x + \sin x\tan x = \sec x$

Answer 1

Hint: Here we just need to apply the general formula of $\tan x = \dfrac{{\sin x}}{{\cos x}}$ and then we can simplify it further to get the required result. We can also use the property that ${\sin ^2}x + {\cos ^2}x = 1$ to get the required proof.

Complete step by step solution:
Here we are given to prove that $\cos x + \sin x\tan x = \sec x$
So let us start to solve for the left hand side of the proof and proceed till we prove it. So we know that $\tan x = \dfrac{{\sin x}}{{\cos x}}$
And we are given $\cos x + \sin x\tan x$$ - - - - (1)$
Let us put the value of $\tan x$ in equation (1) and we will get that:
$\cos x + \sin x\left( {\dfrac{{\sin x}}{{\cos x}}} \right) = \sec x$$ - - - - (2)$
So we can simplify it and we will get:
$\cos x + \left( {\dfrac{{{{\sin }^2}x}}{{\cos x}}} \right) = \sec x$
Now let us take the LCM as $\cos x$ and we will get:
$\left( {\dfrac{{{{\sin }^2}x + {{\cos }^2}x}}{{\cos x}}} \right)$$ - - - - (3)$
Now we know that in the numerator we have got the term which is ${\sin ^2}x + {\cos ^2}x$ and we know that in the trigonometric function we have the property by which we know that
${\sin ^2}x + {\cos ^2}x = 1$
So let us substitute this value of ${\sin ^2}x + {\cos ^2}x$ in equation (3) and we will get:
$\left( {\dfrac{{{{\sin }^2}x + {{\cos }^2}x}}{{\cos x}}} \right) = \left( {\dfrac{1}{{\cos x}}} \right)$
Now we know that every trigonometric function has its reciprocal like reciprocal of sin is cosec and therefore we will get:
$\left( {\dfrac{{{{\sin }^2}x + {{\cos }^2}x}}{{\cos x}}} \right) = \left( {\dfrac{1}{{\cos x}}} \right) = \sec x$
This is because we know that reciprocal of $\cos x$ is $\sec x$.

Note:
We can proceed with these types of problems in many ways. Another way can be by taking the $\cos x$ common and we will get:
$\cos x + \sin x\tan x = \sec x$
$\cos x\left( {1 + \dfrac{{\sin x}}{{\cos x}}\tan x} \right) = \sec x$
Now we can say that $\tan x = \dfrac{{\sin x}}{{\cos x}}$ and we will get:
$\cos x\left( {1 + {{\tan }^2}x} \right) = \sec x$ and we know that $\left( {1 + {{\tan }^2}x} \right) = {\sec ^2}x = \dfrac{1}{{{{\cos }^2}x}}$
Hence we will get:
$\cos x\left( {1 + {{\tan }^2}x} \right) = \sec x$
$\cos x\left( {\dfrac{1}{{{{\cos }^2}x}}} \right) = \dfrac{1}{{\cos x}} = \sec x$