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How do you prove the identity $\dfrac{{\tan x}}{{\cos ecx + \cot x}} = \sec x - 1$ ?

Last updated date: 23rd Feb 2024
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Hint: To solve such questions start by separately computing the left-hand side and right-hand side of the given equation. Next, write each term in the given equation to its corresponding $\sin x$ and $\cos x$ values. Then check whether the left-hand side and right-hand side are equal.

Given $\dfrac{{\tan x}}{{\cos ecx + \cot x}} = \sec x - 1$
First consider the left-hand side of the given equation, that is,
$\dfrac{{\tan x}}{{\cos ecx + \cot x}}......(1)$
We know that
$\tan x = \dfrac{{\sin x}}{{\cos x}}$
Also, it is known that
$\cos ecx = \dfrac{1}{{\sin x}}$
And
$\cot x = \dfrac{{\cos x}}{{\sin x}}$
Substitute these in the equation $\left( 1 \right)$ , that is,
$\dfrac{{\dfrac{{\sin x}}{{\cos x}}}}{{\dfrac{1}{{\sin x}} + \dfrac{{\cos x}}{{\sin x}}}}$
Rearranging the terms we get,
$\dfrac{{\sin x}}{{\cos x\left( {\dfrac{{1 + \cos x}}{{\sin x}}} \right)}}$
Taking $\sin x$ to the numerator we get
$\dfrac{{{{\sin }^2}x}}{{\cos x\left( {1 + \cos x} \right)}}......(2)$
We also know that,
${\sin ^2}x + {\cos ^2}x = 1$
Rearranging this we get,
${\sin ^2}x = 1 - {\cos ^2}x$
Substituting this in the equation $\left( 2 \right)$
$\dfrac{{1 - {{\cos }^2}x}}{{\cos x\left( {1 + \cos x} \right)}}$
We know the identity ${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$ , so we can write the above equation as,
$\dfrac{{\left( {1 + \cos x} \right)\left( {1 - \cos x} \right)}}{{\cos x\left( {1 + \cos x} \right)}}$
Canceling out the common terms in the numerator and denominator, we get,
$\dfrac{{1 - \cos x}}{{\cos x}}......(3)$
Next consider the right-hand side of $\dfrac{{\tan x}}{{\cos ecx + \cot x}} = \sec x - 1$ , that is
$\sec x - 1$
We know that,
$\sec x = \dfrac{1}{{\cos x}}$
Substituting this on the right-hand side, we get,
$\dfrac{1}{{\cos x}} - 1$
Further simplifying we get,
$\dfrac{{1 - \cos x}}{{\cos x}}......(4)$
From equation $\left( 3 \right)$ and equation $\left( 4 \right)$ , it can be seen that LHS is equal to RHS.
Therefore, proved the identity $\dfrac{{\tan x}}{{\cos ecx + \cot x}} = \sec x - 1$ .

All trigonometric ratios of angle can be easily found if one of the trigonometric ratios of an acute angle is given. The value of $\sin x$ or $\cos x$ never goes beyond $1$ . Also, the value of $\sec x$ or $\cos ecx$ is always greater than or equal to $1$ .
Note: In order to solve these types of questions, one should be very clear about the trigonometric identities. The simplest way to solve this type of question is to equate each term of the equation to its corresponding $\sin x$ and $\cos x$ values