Question

tanA/1+cotA+CotA/1-tanA=1+secAcosecA

Answer 1

Hey there !!

→ Prove that :-)

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

→ Solution :-)

[Let A = $\theta$ ] .

Given,,

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

$= 1 + \frac{ \sin \theta}{ \cos \theta} + \frac{ \cos \theta}{ \sin \theta} . \\ \\ = \frac{cos \theta \sin \theta + { \sin}^{2} \theta + { \cos}^{2} \theta}{ \cos \theta \sin \theta} . \\ \\ = \frac{1 + \cos \theta \sin \theta}{ \cos \theta \sin \theta}. \\ \\ = \frac{1}{ \cos \theta \sin \theta} + \frac{ \cancel{\cos \theta \sin \theta}}{ \cancel{ \cos \theta \sin \theta}} . \\ \\ = \frac{1}{ \cos \theta} \times \frac{1}{ \sin \theta} + 1. \\ \\ = 1 + \sec \theta cosec \theta . \checkmark \checkmark$

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

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

✔✔ Hence, it is proved ✅ ✅.

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

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

Answer 2

Hlo mate :-

Solution :-

● See the attached pic

☆ ☆ ☆ Hop It's helpful ☆ ☆ ☆