Question

How do you simplify the following expression to a single trigonometric function $\dfrac{{\sin x}}{{1 - \cos x}} - \cos ecx$ ?

Answer 1

In order to solve this equation we will use a simple approach. We will convert the cosec function to sin and then we will take the LCM. After this we will be using the fundamental identity of trigonometric functions to find the single trigonometric function.

Complete step by step solution:
Given that $\dfrac{{\sin x}}{{1 - \cos x}} - \cos ecx$.We know that $\cos ecx$ can be written as $\dfrac{1}{{\sin x}}$. So rewriting the equation,
$\dfrac{{\sin x}}{{1 - \cos x}} - \dfrac{1}{{\sin x}}$
Taking the LCM we get,
$\dfrac{{\sin x.\sin x - \left( {1 - \cos x} \right)}}{{\sin x\left( {1 - \cos x} \right)}}$
On multiplying the sin terms we get,
$\dfrac{{{{\sin }^2}x - \left( {1 - \cos x} \right)}}{{\sin x\left( {1 - \cos x} \right)}}$
On multiplying the minus with bracket,
$\dfrac{{{{\sin }^2}x - 1 + \cos x}}{{\sin x\left( {1 - \cos x} \right)}}$
We know that ${\sin ^2}x + {\cos ^2}x = 1$ can be written as, ${\sin ^2}x - 1 = - {\cos ^2}x$

So we can write the above equation as ,
$\dfrac{{ - {{\cos }^2}x + \cos x}}{{\sin x\left( {1 - \cos x} \right)}}$
On rearranging the terms
$\dfrac{{\cos x - {{\cos }^2}x}}{{\sin x\left( {1 - \cos x} \right)}}$
Taking cos function common,
$\dfrac{{\cos x\left( {1 - \cos x} \right)}}{{\sin x\left( {1 - \cos x} \right)}}$
Cancelling the common bracket,
$\dfrac{{\cos x}}{{\sin x}}$
We know that $\cot x = \dfrac{{\cos x}}{{\sin x}}$

Thus the final answer is $\dfrac{{\sin x}}{{1 - \cos x}} - \cos ecx = \cot x$.

Note: Here note that cosec function is the indicator to proceed towards the solution because the numerator of first terms is a sin function that is reciprocal of cosec function. We have to convert the given expression to a single function so try to write the ratios or terms nearer to a same function or such that we can use the different trigonometric identities to get the single function.