Question

How do you verify $\dfrac{{\cos 2x - 1}}{{\sin 2x}} = - \tan x$

Answer 1

Hint: We will begin with the left-hand side of the equation and then we will solve it by using the formula $\cos 2x = 1 - 2{\sin ^2}x$ in the numerator and $\sin 2x = 2\sin x\cos x$ in the denominator. After that cancel out the common factor from the numerator and denominator. After that take the right-hand side and convert all terms in sin and cos. Then equate both sides to verify the terms.

Complete step-by-step answer:
It is mentioned in the question that
$ \Rightarrow \dfrac{{\cos 2x - 1}}{{\sin 2x}} = - \tan x$ ….. (1)
Now beginning with the left-hand side of the equation (1) we get,
$ \Rightarrow \dfrac{{\cos 2x - 1}}{{\sin 2x}}$
We know that, $\cos 2x = 1 - 2{\sin ^2}x$ and $\sin 2x = 2\sin x\cos x$.
Substitute these values in the numerator and denominator of the above expression
$ \Rightarrow \dfrac{{1 - 2{{\sin }^2}x - 1}}{{2\sin x\cos x}}$
Subtract the like terms in the numerator,
$ \Rightarrow \dfrac{{ - 2{{\sin }^2}x}}{{2\sin x\cos x}}$
Cancel out the common factor from the numerator and denominator,
$ \Rightarrow - \dfrac{{\sin x}}{{\cos x}}$ ….. (2)
Now take the right-hand side of the equation (1) we get,
$ \Rightarrow - \tan x$
Now converting tan in the above equation in terms of sin and cos and hence we get,
$ \Rightarrow - \dfrac{{\sin x}}{{\cos x}}$ ….. (3)
Since from equation (2), we can say that the left-hand side is equal to the right-hand side in equation (3).

Hence, we have verified the given expression.

Note:
 In trigonometry remembering the formulas and the identities is very important because then it becomes easy. We may get confused about how to proceed further but here the key is to substitute the formula of $\cos 2x$ in terms of the sine function. Then we need to change $\sin 2x$ also. After that cancel out terms from the numerator and the denominator to get the left-hand side equal to the right-hand side of the equation.