Question

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

Answer 1

We use trigonometric identities and ratios to solve this problem. We use some methods of simplifying algebraic expressions and verify the equation by equating the left-hand side and right-hand side of the given equation.
The formulas that we use in this problem are ${\cos ^2}x + {\sin ^2}x = 1$ and ${\sec ^2}x - {\tan ^2}x = 1$ which are standard trigonometric identities.

Complete step by step answer:
To solve this problem, consider the left-hand side of the equation and the right-hand side of the equation. And then we simplify the left-hand side of the equation and check whether we are getting the same expression present on the right-hand side as our result.
Now, in left hand side of equation, it is given as $\dfrac{{\tan x - \sec x + 1}}{{\tan x + \sec x - 1}}$
We know the identity ${\sec ^2}x - {\tan ^2}x = 1$
So, substitute the value of 1, in the denominator.
$\Rightarrow \dfrac{{\tan x - \sec x + 1}}{{\tan x + \sec x - ({{\sec }^2}x - {{\tan }^2}x)}}$
We all know that, ${a^2} - {b^2} = (a + b)(a - b)$
So, we can write ${\sec ^2}x - {\tan ^2}x = (\sec x + \tan x)(\sec x - \tan x)$
So, we get,
$\Rightarrow \dfrac{{\tan x - \sec x + 1}}{{\tan x + \sec x - (\sec x + \tan x)(\sec x - \tan x)}}$
Now take the term $\sec x + \tan x$ common in the denominator.
$\Rightarrow \dfrac{{\tan x - \sec x + 1}}{{(\tan x + \sec x)(1 - (\sec x - \tan x))}}$
$\Rightarrow \dfrac{{\tan x - \sec x + 1}}{{(\tan x + \sec x)(1 - \sec x + \tan x)}}$
Now, you can observe that, there is a term common in numerator and denominator which is $\tan x - \sec x + 1$, so they get cancelled off.
$\Rightarrow \dfrac{1}{{\tan x + \sec x}}$
We know that, $\tan x = \dfrac{{\sin x}}{{\cos x}}$ and $\sec x = \dfrac{1}{{\cos x}}$.
So, substitute these values in the expression.
$\Rightarrow \dfrac{1}{{\dfrac{{\sin x}}{{\cos x}} + \dfrac{1}{{\cos x}}}}$
$\Rightarrow \dfrac{1}{{\dfrac{{\sin x + 1}}{{\cos x}}}}$
And finally, we can conclude that,
$\Rightarrow \dfrac{{\cos x}}{{1 + \sin x}} = RHS$

Note:
There is another way of verifying this equation.
We can also solve this problem in another way by multiplying both the numerator and denominator by ${\cos ^2}x$.
$LHS \Rightarrow \dfrac{{{{\cos }^2}x(\tan x - \sec x + 1)}}{{{{\cos }^2}x(\tan x + \sec x - 1)}}$
Take a $\cos x$ from ${\cos ^2}x$ in the numerator and multiply it to the next term.
$LHS \Rightarrow \dfrac{{\cos x(\tan x.\cos x - \sec x.\cos x + 1.\cos x)}}{{\tan x.{{\cos }^2}x + \sec x.{{\cos }^2}x - 1.{{\cos }^2}x}}$
Now, we know that, $\tan x = \dfrac{{\sin x}}{{\cos x}}$ and $\sec x = \dfrac{1}{{\cos x}}$.
So, substitute these values in the expression.
$\Rightarrow \dfrac{{\cos x\left( {\dfrac{{\sin x}}{{\cos x}}.\cos x - \dfrac{1}{{\cos x}}.\cos x + 1.\cos x} \right)}}{{\left( {\dfrac{{\sin x}}{{\cos x}}.{{\cos }^2}x + \dfrac{1}{{\cos x}}.{{\cos }^2}x - 1.{{\cos }^2}x} \right)}}$
$\Rightarrow \dfrac{{\cos x\left( {\sin x - 1 + \cos x} \right)}}{{\left( {\sin x.\cos x + \cos x - {{\cos }^2}x} \right)}}$
We know that, ${\cos ^2}x = 1 - {\sin ^2}x$
$\Rightarrow \dfrac{{\cos x\left( {\sin x - 1 + \cos x} \right)}}{{\sin x.\cos x + \cos x - (1 - {{\sin }^2}x)}}$
As we know the formula ${a^2} - {b^2} = (a + b)(a - b)$
$\Rightarrow \dfrac{{\cos x\left( {\sin x - 1 + \cos x} \right)}}{{\cos x(1 + \sin x) - (1 - \sin x)(1 + \sin x)}}$
$\Rightarrow \dfrac{{\cos x\left( {\sin x - 1 + \cos x} \right)}}{{(1 + \sin x)(\cos x - (1 - \sin x))}}$
So, we can conclude that,
$LHS \Rightarrow \dfrac{{\cos x\left( {\sin x - 1 + \cos x} \right)}}{{(1 + \sin x)(\cos x - 1 + \sin x)}}$
$LHS \Rightarrow \dfrac{{\cos x}}{{1 + \sin x}}$
So, here, LHS is equal to RHS.