Question

tan³θ-1/tanθ-1=sec²θ+tanθ Prove it?

Answer 1

Steps:

1) We know , 

$( x^{3} -y^{3}) = (x-y)(x^{2}+y^{2}+xy)$ 

2)  LHS= 

Put

$x=\tan(\theta) \\y = 1$

on the above cubic factorization .

Assuming  

$\tan(\theta)\neq 1$

to avoid danger with Indeterminate forms;

$\frac{tan^{3}(\theta)-1}{tan(\theta)-1} =  \frac{(tan(\theta)-1)(tan^{2}(\theta)+1^{2}+tan(\theta)*1)}{tan(\theta)-1} \\ \\ =\ \textgreater \  tan^{2}(\theta)+1^{2}+tan(\theta) \\ \\ =\ \textgreater \  sec^{2}(\theta)+tan(\theta)$

Hence Proved:  

$\frac{\tan^3(\theta)-1}{\tan(\theta)-1}=sec^2(\theta)+\tan(\theta)$ for

$tan(\theta)\neq 1$

Answer 2

$LHS$ = (tan³θ - 1)/(tanθ - 1)

We know, a³ - b³ = (a - b)(a² + b² + ab)
So, tan³θ - 1 = (tanθ - 1)(tan²θ + 1 + tanθ)
we know, se²θ - tan²θ = 1 ∴ sec²θ = 1 + tan²θ
∴ [(tan²θ + 1 ) + tanθ] = [sec²θ + tanθ]
so, tan³θ - 1 = (tanθ - 1)(sec²θ + tanθ)

Now, (tan³θ - 1)/(tanθ - 1) = (tanθ - 1)(sec²θ + tanθ)/(tanθ - 1)
= (sec²θ + tanθ) = RHS
$\bf{hence\: proved}$