Question

#TRIGONOMETRY TIME ☺☺
Prove that :-


$\bf{ \frac{tan A}{(1 - cot A)} + \frac{cot A}{(1 - tan A)} = ( 1 + tan A + cot A ) .}$

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Answer 1

$\large \green{ \boxed{ \boxed{ \mathbb{TRIGONOMETRY.}}}}$

→ Prove that :-

$\bf{ \frac{tan A}{(1 - cot A)} + \frac{cot A}{(1 - tan A)} = ( 1 + tan A + cot A ) .}$

Step-by-step explanation :-

Let A = $\theta$

This question can be solve by two methods.

1st method :-

$\orange {\boxed{ \tt See \: attachment }}$

2nd method :-

→ Solution :-)

Given,,

$\begin{lgathered}\begin{lgathered}\frac{\tan\theta}{1-\cot\theta}\;+\;\frac{\cot\theta}{1-\tan\theta}\\ \\=\frac{\tan\theta}{1-\cot\theta}\;+\;\frac{\cot\theta}{1-\frac{1}{\cot\theta}}\\ \\=\frac{\tan\theta}{1-\cot\theta}+\frac{\cot^{2}\theta}{\cot\theta-1}\\ \\=\frac{\tan\theta}{1-\cot\theta}-\frac{\cot^{2}\theta}{1-\cot\theta}\\ \\=\frac{\tan\theta-\cot^{2}\theta}{1-\cot\theta}\\ \\ =\frac{\frac{1}{\cot\theta}-\cot^{2}\theta}{1-\cot\theta}\\ \\=\frac{1-cot^{3}\theta}{\cot\theta(1-\cot\theta)}\\ \\=\frac{(1-cot\theta)(1+cot^{2}\theta+\cot\theta)}{\cot\theta(1-\cot\theta)}\\ \\=\frac{1+cot^{2}\theta+\cot\theta}{\cot\theta}\\ \\=\frac{1}{\cot\theta}+\frac{\cot^{2}\theta}{\cot\theta}+\frac{\cot\theta}{\cot\theta}\\ \\=\tan\theta+\cot\theta+1\;\;\;\textbf{Proved.}\end{lgathered}\end{lgathered}$

$\huge \pink{ \bf \underline{ \underline \mathbb{LHS = RHS.}}}$

Hence, it is proved .

Answer 2

Trigonometry :

Explanation :

Refer the attached picture.