Question

How do you prove $\left( {\dfrac{{\tan x}}{{1 - \cot x}}} \right) + \left( {\dfrac{{\cot x}}{{1 - \tan x}}} \right) = 1 + \sec x\cos ecx$ ?

Answer 1

In this question, we have been asked to prove that LHS of the given equation is equal to its RHS. The given equation is a trigonometric equation. Start with LHS as I do not see much scope in RHS. Convert the ratios of tan and cot into sin and cos. Take LCM in the denominator and, using that, simplify the denominator. Then, flip the denominators and multiply them with the numerators. Take minus common from the denominator of the second term. Then, again take the LCM and simplify the terms. You will see the cubes in the numerator. Use the identity of cubes and solve it. Cancel the like terms and then again simplify using trigonometric ratios.

Complete step-by-step solution:
We are given a trigonometric equation. Let us start solving it –
Now we will convert the identities into sin and cos as I feel easier and comfortable with these ratios.
LHS –
$\Rightarrow \left( {\dfrac{{\tan x}}{{1 - \cot x}}} \right) + \left( {\dfrac{{\cot x}}{{1 - \tan x}}} \right)$
Using$\tan x = \dfrac{{\sin x}}{{\cos x}},\cot x = \dfrac{{\cos x}}{{\sin x}}$,
$\Rightarrow \left( {\dfrac{{\dfrac{{\sin x}}{{\cos x}}}}{{1 - \dfrac{{\cos x}}{{\sin x}}}}} \right) + \left( {\dfrac{{\dfrac{{\cos x}}{{\sin x}}}}{{1 - \dfrac{{\sin x}}{{\cos x}}}}} \right)$
Taking LCM in the denominator,
$\Rightarrow \left( {\dfrac{{\dfrac{{\sin x}}{{\cos x}}}}{{\dfrac{{\sin x}}{{\sin x}} - \dfrac{{\cos x}}{{\sin x}}}}} \right) + \left( {\dfrac{{\dfrac{{\cos x}}{{\sin x}}}}{{\dfrac{{\cos x}}{{\cos x}} - \dfrac{{\sin x}}{{\cos x}}}}} \right)$
Now, we will flip the denominators and multiply them,
$\Rightarrow \left( {\dfrac{{{{\sin }^2}x}}{{\cos x\left( {\sin x - \cos x} \right)}}} \right) + \left( {\dfrac{{{{\cos }^2}x}}{{\sin x\left( {\cos x - \sin x} \right)}}} \right)$
Taking minus common from the denominator (second term),
$\Rightarrow \left( {\dfrac{{{{\sin }^2}x}}{{\cos x\left( {\sin x - \cos x} \right)}}} \right) - \left( {\dfrac{{{{\cos }^2}x}}{{\sin x\left( {\sin x - \cos x} \right)}}} \right)$
Taking LCM in the denominator,
$\Rightarrow \dfrac{{{{\sin }^2}x \times \sin x}}{{\sin x\cos x\left( {\sin x - \cos x} \right)}} - \dfrac{{{{\cos }^2}x \times \cos x}}{{\sin x\cos x\left( {\sin x - \cos x} \right)}}$
Simplifying,
$\Rightarrow \dfrac{{{{\sin }^3}x - {{\cos }^3}x}}{{\sin x\cos x\left( {\sin x - \cos x} \right)}}$
Using ${a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right)$ in numerator,
$\Rightarrow \dfrac{{\left( {\sin x - \cos x} \right)\left( {{{\sin }^2}x + \sin x\cos x + {{\cos }^2}x} \right)}}{{\sin x\cos x\left( {\sin x - \cos x} \right)}}$
On simplifying,
$\Rightarrow \dfrac{{\left( {{{\sin }^2}x + \sin x\cos x + {{\cos }^2}x} \right)}}{{\sin x\cos x}}$
Using the identity ${\sin ^2}x + {\cos ^2}x = 1$,
$\Rightarrow \dfrac{{\left( {1 + \sin x\cos x} \right)}}{{\sin x\cos x}}$
Separating the terms,
$\Rightarrow \dfrac{1}{{\sin x\cos x}} + \dfrac{{\sin x\cos x}}{{\sin x\cos x}}$
Using $\dfrac{1}{{\sin x}} = \cos ecx,\dfrac{1}{{\cos x}} = \sec x$,
$\Rightarrow \sec x\cos ecx + 1$
Therefore, LHS = RHS.
Hence proved.

Note: We can also solve the question without converting the ratios into sin and cos. In such a case, we start by taking LCM. Then, open the brackets and multiply the terms. Use some identities in terms of cot and tan and then, simplify. You will get your answer. There is no such rule which says that we have to expand or start the question from LHS. We can also start from RHS. But always prefer that side which has more scope of expanding.