Question

Prove that $\cos ec2x + \cot 2x = \cot x$?

Answer 1

Hint:The given question deals with basic simplification of trigonometric functions by using some of the simple trigonometric formulae such as $\cot x = \dfrac{{\cos x}}{{\sin x}}$ and $\tan x = \dfrac{{\sin x}}{{\cos x}}$. Then, we use the double angle formulae of sine and cosine. Basic algebraic rules and trigonometric identities are to be kept in mind while simplifying the given problem and proving the result given to us.

Complete step by step answer:
In the given problem, we have to prove a trigonometric equality. Now, we need to make the left and right sides of the equation equal.
$L.H.S. = \cos ec2x + \cot 2x$
So, we use the trigonometric formulae $\cot x = \dfrac{{\cos x}}{{\sin x}}$ and $\tan x = \dfrac{{\sin x}}{{\cos x}}$.
$L.H.S. = \dfrac{1}{{\sin 2x}} + \dfrac{{\cos 2x}}{{\sin 2x}}$
Since the denominators are the same. We can directly add the numerators. So, we get,
$L.H.S. = \dfrac{{1 + \cos 2x}}{{\sin 2x}}$

Now, we make use of the double angle formula of cosine $\cos 2x = 2{\cos ^2}x - 1$ in the numerator and double angle formula of sine $\sin 2x = 2\sin x\cos x$ in the denominator. So, we get,
$L.H.S. = \dfrac{{1 + 2{{\cos }^2}x - 1}}{{2\sin x\cos x}}$
Cancelling like terms with opposite signs,
$L.H.S. = \dfrac{{2{{\cos }^2}x}}{{2\sin x\cos x}}$
Cancelling common factors in numerator and denominator,
$L.H.S. = \dfrac{{\cos x}}{{\sin x}}$
Using the trigonometric formula $\cot x = \dfrac{{\cos x}}{{\sin x}}$, we get,
$L.H.S. = \cot x$
Now, R.H.S $ = \cot x$
As the left side of the equation is equal to the right side of the equation, we have,
$\cos ec2x + \cot 2x = \cot x$
Hence, Proved.

Note:Besides these simple trigonometric formulae, trigonometric identities are also of significant use in such type of questions where we have to simplify trigonometric expressions with help of basic knowledge of algebraic rules and operations. There are various forms of double angle formulae of cosine such as $\cos 2x = 2{\cos ^2}x - 1$, $\cos 2x = 1 - 2{\sin ^2}x$ and $\cos 2x = \dfrac{{1 - {{\tan }^2}x}}{{1 + {{\tan }^2}x}}$. We use the formula $\cos 2x = 2{\cos ^2}x - 1$ because it helps us simplify the numerator of the function easily.