Question

Tan³x‐1/tan x‐1=sec²x+ tanx

Answer 1

We have to prove that,

$\dfrac{\tan^3(x)-1}{\tan(x)-1}=\sec^2(x)+\tan(x)$

Here we may recall the identity,

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

By using this,

$\tan^3(x)-1=\tan^3(x)-1^3=(\tan(x)-1)(\tan^2(x)+\tan(x)+1)$

Also we use,

$\sec^2(x)-\tan^2(x)=1\ \ \ \ \ \Longrightarrow\ \ \ \ \ \sec^2(x)=\tan^2(x)+1$

So,

$\begin{aligned}&\large\textsf{LHS}\\ \\ \Longrightarrow\ \ &\frac{\tan^3(x)-1}{\tan(x)-1}\\ \\ \Longrightarrow\ \ &\frac{(\tan(x)-1)(\tan^2(x)+\tan(x)+1)}{(\tan(x)-1)}\\ \\ \Longrightarrow\ \ &\tan^2(x)+\tan(x)+1\\ \\ \Longrightarrow\ \ &\tan^2(x)+1+\tan(x)\\ \\ \Longrightarrow\ \ &\sec^2(x)+\tan(x)\\ \\ \Longrightarrow\ \ &\large\textsf{RHS}\end{aligned}$

Hence Proved!

One thing we have to keep in mind here is that tan(x) should not be equal to 1.

Because, if tan(x) = 1, then tan(x) - 1 = 0, thus the LHS will be an indeterminate form, since the denominator in the LHS is tan(x) - 1.

tan(x) will be equal to 1 if and only if 'x' is in the form of  (π/4) + nπ = [(4n + 1)/4]π  radian, ∀n ∈ Z.

Thus this proof is only applicable for every x ≠ [(4n + 1)/4]π  rad.

Answer 2

Answer:-

Refer the attachment