Question

Tan3-1/tan-1=sec2+tan
prove it​

Answer 1

Step-by-step explanation:

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

LHS

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

                                                                                           $sin^{2}θ - tan^{2} θ = 1\\\\sin^{2}θ = 1 +tan^{2} θ$

$= (sec^{2}θ + tanθ)$

RHS

= $(sec^{2}θ + tanθ)$

Therefore LHS = RHS

                               HENCE PROVED