Given, tan ^3θ 1tanθ 1= ^2θ +tanθ LHS: =tan ^3θ 1tanθ 1 using a^3 b^3 formula, we get, =(tanθ 1)(tan ^2θ +tanθ× 1+1)tanθ 1 =tan ^2θ ...

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General Solution of Trigonometric Equation

tan3θ-1tanθ-1...

Question

$\frac{{tan}^{3} θ - 1}{tan θ - 1} = {sec}^{2} θ + tan θ$

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Solution

Given,

$\frac{{tan}^{3} θ - 1}{tan θ - 1} = {sec}^{2} θ + tan θ$

LHS:

$= \frac{{tan}^{3} θ - 1}{tan θ - 1}$

using $a^{3} - b^{3}$ formula, we get,

$= \frac{(tan θ - 1) ({tan}^{2} θ + tan θ \times 1 + 1)}{tan θ - 1}$

$= {tan}^{2} θ + tan θ + 1$

$= {sec}^{2} θ + tan θ$

:RHS

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