Question

How do you verify $\dfrac{{\left( {\cos 2x - 1} \right)}}{{\sin 2x}} = - \tan x$?

Answer 1

Hint: In order to solve this, we need to solve the right hand side and get an answer which is equal to the left hand side to prove and verify our sum.
We know that $\sin 2x = 2\sin x\cos x$ and $\cos 2x = 1 - 2{\sin ^2}x$.
We substitute the values in the given equation and solve it to get our required answer and verify the sum by proving the right hand side to be equal to the left hand side.

Formula used: $\sin 2x = 2\sin x\cos x$ and $\cos 2x = 1 - 2{\sin ^2}x$

Complete step-by-step solution:
In the given question we are asked to verify a trigonometric function by making the right hand side equal to the left hand side.
In the right hand side, we have: $\dfrac{{\cos 2x - 1}}{{\sin 2x}}$ ,
Now we need to solve it in such a way so that the answer is equal to the left hand side which is $ - \tan x$
Let us solve the right hand side first:
$ \Rightarrow \dfrac{{\cos 2x - 1}}{{\sin 2x}}$
We know that: $\sin 2x = 2\sin x\cos x$ and $\cos 2x = 1 - 2{\sin ^2}x$ , placing these values in the above expression, we get:
$ \Rightarrow \dfrac{{\left( {1 - 2{{\sin }^2}x} \right) - 1}}{{2\sin x\cos x}}$
On simplifying it further, we get:
$ \Rightarrow \dfrac{{ - 2{{\sin }^2}x}}{{2\sin x\cos x}}$
\[ \Rightarrow \dfrac{{ - \not{2}{{\sin }^{\not{2}}}x}}{{\not{2}\not{{\sin x}}\cos x}}\]
On simplifying, we get:
$ \Rightarrow \dfrac{{ - \sin x}}{{\cos x}}$
Now we know that $\dfrac{{\sin x}}{{\cos x}} = \tan x$
Therefore, $\dfrac{{ - \sin x}}{{\cos x}} = - \tan x$ , which is equal to the left hand side.
Hence proved.

Note: Trigonometry is a branch of mathematics which deals with triangles. There are many trigonometric formulas that establish a relation between the lengths and angles of respective triangles. In trigonometry, we use a right-angled triangle to find ratios of its different sides and angles such as sine, cosine, tan, and their respective inverse like cosec, sec, and cot. Some common formulas of trigonometric identities are:
${{sin\theta = }}\dfrac{{{\text{perpendicular}}}}{{{\text{hypotenuse}}}}$ , where perpendicular is the side containing the right angle in a right angled triangle and hypotenuse is the side opposite to the perpendicular.
${{cos\theta = }}\dfrac{{{\text{base}}}}{{{\text{hypotenuse}}}}$ , where base is the side containing the perpendicular and hypotenuse
${{tan\theta = }}\dfrac{{{\text{perpendicular}}}}{{{\text{base}}}}$