Question

Prove
$\frac{ \tan \theta }{1 - cot \theta} + \frac{cot \theta}{1 - tan \theta} = 1 + tan \theta + cot \theta$
​

Answer 1

Given :-

$\sf\dfrac{\tan\theta}{1-\cot\theta}+\dfrac{\cot\theta}{1-\tan\theta}=1+\tan\theta+\cot\theta$

To Find :-

We need to prove the given equation

Solution :-

In these type of question we need to separate the LHS and RHS

LHS

$\sf\dfrac{\tan\theta}{1-\cot\theta}+\dfrac{\cot\theta}{1-\tan\theta}$

We know that

$\bullet \tan\theta=\dfrac{1}{\cot\theta}$

$\sf\dfrac{\tan\theta}{1-\cot\theta}+\dfrac{\cot\theta}{1-\dfrac{1}{\cot\theta}}$

$\sf\dfrac{\tan\theta}{1-\cot\theta}+\dfrac{\cot^2\theta}{1-\cot\theta}$

$\sf\dfrac{\tan\theta+\cot^2\theta}{cot\theta}$

$\sf\dfrac{\dfrac{1}{\cot\theta}-\cot^2\theta}{1-\cot\theta}$

$\sf \dfrac{1-\cot^3\theta}{\cot\theta(1-\cot\theta)}$

$\sf \bullet (a-b)^3 = (a-b)(a^2+b^2-2ab)$

$\sf\dfrac{(1-\cot\theta)(1+\cos^2\theta+\cot\theta)}{\cot\theta(1-\cot\theta)}$

$\sf\dfrac{1}{\cot\theta}+\dfrac{\cot^2\theta}{\cot\theta}+\dfrac{\cot\theta}{\cot\theta}$

$\sf \tan\theta+\cot\theta+1$

$\sf 1+\tan\theta+\cot\theta$

RHS

$\sf 1+\tan\theta+\cot\theta$

Answer 2

$\large\underline{\sf{Solution-}}$

Consider, LHS

$\rm :\longmapsto\:\dfrac{tan\theta }{1 - cot\theta } + \dfrac{cot\theta }{1 - tan\theta }$

We know,

$\boxed{ \tt{ \: cotx = \frac{1}{tanx} \: }}$

So, using this identity, we get

$\rm \:  =  \:\dfrac{tan\theta }{1 - \dfrac{1}{tan\theta } } + \dfrac{\dfrac{1}{tan\theta } }{1 - tan\theta }$

$\rm \:  =  \:\dfrac{tan\theta }{ \dfrac{tan\theta - 1}{tan\theta } } + \dfrac{1 }{tan\theta (1 - tan\theta) }$

$\rm \:  =  \:\dfrac{ {tan}^{2} \theta }{tan\theta - 1} - \dfrac{1}{tan\theta (tan\theta - 1)}$

$\rm \:  =  \:\dfrac{ {tan}^{3} \theta - 1}{tan\theta (tan\theta - 1)}$

We know,

$\boxed{ \tt{ \: {x}^{3} - {y}^{3} = (x - y)( {x}^{2} + xy + {y}^{2}) \: }}$

So, using this identity, we get

$\rm \:  =  \:\dfrac{(tan\theta - 1)( { {tan}^{2} \theta + tan\theta + 1)}}{tan\theta (tan\theta - 1)}$

$\rm \:  =  \:\dfrac{{tan}^{2} \theta + tan\theta + 1}{tan\theta }$

$\rm \:  =  \:\dfrac{ {tan}^{2} \theta }{tan\theta } + \dfrac{tan\theta }{tan\theta } + \dfrac{1}{tan\theta }$

$\rm \:  =  \:tan\theta + 1 + cot\theta$

$\rm \:  =  \:1 + tan\theta + cot\theta$

Hence,

$\boxed{ \tt{ \: \:\dfrac{tan\theta }{1 - cot\theta } + \dfrac{cot\theta }{1 - tan\theta } = 1 + tan\theta + cot\theta \: \: }}$

Additional Information:-

Relationship between sides and T ratios

sin θ = Opposite Side/Hypotenuse

cos θ = Adjacent Side/Hypotenuse

tan θ = Opposite Side/Adjacent Side

sec θ = Hypotenuse/Adjacent Side

cosec θ = Hypotenuse/Opposite Side

cot θ = Adjacent Side/Opposite Side

Reciprocal Identities

cosec θ = 1/sin θ

sec θ = 1/cos θ

cot θ = 1/tan θ

sin θ = 1/cosec θ

cos θ = 1/sec θ

tan θ = 1/cot θ

Co-function Identities

sin (90°−x) = cos x

cos (90°−x) = sin x

tan (90°−x) = cot x

cot (90°−x) = tan x

sec (90°−x) = cosec x

cosec (90°−x) = sec x

Fundamental Trigonometric Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

cosec²θ - cot²θ = 1