Question

How do you prove that: $\tan \dfrac{x}{2} = \cos ecx - \cot x$ ?

Answer 1

The given question deals with proving a trigonometric equality using the basic and simple trigonometric formulae and identities such as $\cos ecx = \dfrac{1}{{\sin x}}$ and $\cot x = \dfrac{{\cos x}}{{\sin x}}$. We first convert all the trigonometric functions into sine and cosine in order to simplify the expression using basic algebraic identities and rules.

Complete step by step answer:
Now, we need to make the left and right sides of the equation equal.
R.H.S. $= \cos ecx - \cot x$
So, we will convert all the trigonometric functions into sine and cosine using trigonometric formulae and identities. So, using the trigonometric formula $\cos ecx = \dfrac{1}{{\sin x}}$ and $\cot x = \dfrac{{\cos x}}{{\sin x}}$, we get, $\dfrac{1}{{\sin x}} - \dfrac{{\cos x}}{{\sin x}}$.Since the denominators of both the rational trigonometric expressions are the same. So, we just add up the numerators directly. Hence, we get, $\dfrac{{1 - \cos x}}{{\sin x}}$.
Now, we know the half angle formula for cosine as $\cos x = 1 - 2{\sin ^2}\dfrac{x}{2}$.
$\dfrac{{1 - \left( {1 - 2{{\sin }^2}\dfrac{x}{2}} \right)}}{{\sin x}}$
Opening the brackets in numerator, we get $\dfrac{{2{{\sin }^2}\dfrac{x}{2}}}{{\sin x}}$.

Using the half angle formula for sine, we get $\dfrac{{2{{\sin }^2}\dfrac{x}{2}}}{{2\sin \dfrac{x}{2}\cos \dfrac{x}{2}}}$.
Cancelling the common terms in numerator and denominator, we get $\dfrac{{\sin \dfrac{x}{2}}}{{\cos \dfrac{x}{2}}}$.
Now, we know that $\tan x = \dfrac{{\sin x}}{{\cos x}}$. Hence, we get,
$\tan \dfrac{x}{2}$
Now, L.H.S $= \tan \dfrac{x}{2}$
As the left side of the equation is equal to the right side of the equation, we have $\tan \dfrac{x}{2} = \cos ecx - \cot x$
Hence, Proved.

Note: Given problem deals with Trigonometric functions. For solving such problems, trigonometric formulae and identities such as $\cot x = \dfrac{{\cos x}}{{\sin x}}$ and the half angle formulae for sine and cosine $\cos x = 1 - 2{\sin ^2}\dfrac{x}{2}$ and $\sin x = 2\sin \dfrac{x}{2}\cos \dfrac{x}{2}$ should be used. We also need knowledge of algebraic rules and identities to simplify the expression. Definitions of the trigonometric functions such as secant $\sec x = \dfrac{1}{{\cos x}}$, cosecant $\cos ecx = \dfrac{1}{{\sin x}}$ and tangent are essential for solving the problem.