Question

show that 1-tan^a/cot^a-1=tan^a ​

Answer 1

Answer:

1-tan^a/cot^a-1

1-tan^a/1/tan^a-1(cot^a=1/tan^a)

1-tan^a/1-tan^a/tan^a(1-tan^a gets cancel

=tan^a

Answer 2

Hey there !!

→ Prove that :-)

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

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

$\ \bf \underline{ \underline \mathsf{LHS = RHS.}}$

Hence, it is proved ____________________________________

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