Question

1+tan^3α/1+tanα+tanα-sec^2α=0​

Answer 1

Step-by-step explanation:

1+tan^3A/1+tanA +tanA-sec^2A (1+tan^3A is only divided by 1+tanA)

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

=(1+tanA)(1-tanA+tan^2A)/(1+tanA) +tanA-sec^2A

=1+tan^2A-tanA+tanA-sec^2A

We know that,

$1+tan^2A=sec^2A$

=sec^2A-sec^2A-tanA+tana

=0

RHS=LHS

0=0(proved)