Question

Prove that:
$\dfrac{{\tan A}}{{(1 - \cot A)}} + \dfrac{{\cot A}}{{(1 - \tan A)}} = \sec A \times \cos ecA + 1$

Answer 1

Hint: Here, we will simplify the L.H.S and convert it into R.H.S of the equation using trigonometric formulae.

Complete step-by-step answer:
$\dfrac{{\tan A}}{{(1 - \cot A)}} + \dfrac{{\cot A}}{{(1 - \tan A)}} = \sec A \times \cos ecA + 1$
Take LHS
As you know $\cot A = \dfrac{1}{{\tan A}}$ , substitute this value
$
  \dfrac{{\tan A}}{{(1 - \dfrac{1}{{\tan A}})}} + \dfrac{{\cot A}}{{(1 - \tan A)}} \\
   \Rightarrow \dfrac{{{{\tan }^2}A}}{{(\tan A - 1)}} + \dfrac{{\cot A}}{{(1 - \tan A)}} \\
   \Rightarrow \dfrac{{{{\tan }^2}A}}{{(\tan A - 1)}} - \dfrac{{\cot A}}{{(\tan A - 1)}} = \dfrac{{{{\tan }^2}A - \dfrac{1}{{\tan A}}}}{{(\tan A - 1)}} = \dfrac{{{{\tan }^3}A - 1}}{{\tan A(\tan A - 1)}} \\
$
In numerator apply ${a^3} - {b^3}$ formula$ = (a - b)({a^2} + {b^2} + ab)$
$ \Rightarrow \dfrac{{(\tan A - 1)({{\tan }^2}A + 1 + \tan A)}}{{\tan A(\tan A - 1)}} = \dfrac{{{{\tan }^2}A + 1 + \tan A}}{{\tan A}}$
As you know ${\tan ^2}A + 1 = {\sec ^2}A$
$ \Rightarrow \dfrac{{{{\sec }^2}A + \tan A}}{{\tan A}} = \dfrac{{{{\sec }^2}A}}{{\tan A}} + 1$
You know $\sec A = \dfrac{1}{{\cos A}},\tan A = \dfrac{{\sin A}}{{\cos A}}$
$ \Rightarrow \dfrac{{{{\sec }^2}A}}{{\tan A}} + 1 = \dfrac{{\cos A}}{{{{\cos }^2}A \times \sin A}} + 1 = \dfrac{1}{{\cos A \times \sin A}} + 1$
You know $\dfrac{1}{{\sin A}} = \cos ecA$
$ \Rightarrow \dfrac{1}{{\cos A \times \sin A}} + 1 = \sec A \times \cos ecA + 1 = R.H.S$

Note: In this type of question always remember the trigonometric formula it will help you in finding your desired answer.