Question

tan A / 1 - cot A + cot A / 1 - tan A = 1+ tan A + cot A

Answer 1

Hey there !!


→ Prove that :-)


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

→ Solution :-)

Given,,

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



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



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



✔✔ Hence, it is proved ✅ ✅.

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$\huge \boxed{ \boxed{ \boxed{ \mathbb{THANKS}}}}$




$\huge \bf{ \# \mathcal{B}e \mathcal{B}rainly.}$

Answer 2

here is your answer OK dude