Question

How do you prove \[cosec x - \cos x \times \cot x = \sin x\]?

Answer 1

Hint:To solve the trigonometric equation in which you have to prove the equation then for that you need to solve both sides of the equation with the known properties of trigonometric functions like summation and subtraction for two or more given quantities.

Complete step by step solution:
The given question is \[cosec x - \cos x \times \cot x = \sin x\]
Here we have to use some transformations, and then after simplifications done to the left hand side of the equation we can achieve the results, the transformation we are going to use are:
\[ \Rightarrow \cos ecx = \dfrac{1}{{\sin x}},\,\cot x = \dfrac{{\cos x}}{{\sin x}}\]
Now using these transformations, we can proceed with the solution, on solving the left hand side of the equation we get:
\[
\Rightarrow \cos ecx - \cos x \times \cot x \\
\Rightarrow \dfrac{1}{{\sin x}} - \cos x \times \dfrac{{\cos x}}{{\sin x}} \\
\Rightarrow \dfrac{{1 - {{\cos }^2}x}}{{\sin x}}(we\,know\,1 - {\cos ^2}x = {\sin ^2}x) \\
\Rightarrow \dfrac{{{{\sin }^2}x}}{{\sin x}} \\
\Rightarrow \sin x \\
\]
Here we see that the left hand part of the equation is equal to the right hand part on solving, hence we can say that both sides of the equation are equal and hence proved.
Formulae Used:
\[ \Rightarrow \cos ecx = \dfrac{1}{{\sin x}},\,\cot x = \dfrac{{\cos x}}{{\sin x}}\]
\[ \Rightarrow \,1 - {\cos ^2}x = {\sin ^2}x\]

Additional Information: Here we can also move further by rearranging the equation that is sending the second term in the left hand side of the question to the right hand side and then after solving the right hand side we can also prove the equation, and it will also satisfy the given equation.

Note: In trigonometric equations we need to solve the equation by using the best possible property suitable for the question, and accordingly we have to proceed keeping in mind that where we have to reach, that is what is asked in question.