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

Hint: 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.