Question

(tanA+2)(2 tanA+1)=5 tanA+sec²A​

Answer 1

Answer:

Step-by-step explanation:

\text{consider,}consider,

(tanA+2)(2tanA+1)(tanA+2)(2tanA+1)

By term by term multiplication we get

=2\,tan^2A+tanA+4\,tanA+2=2tan2A+tanA+4tanA+2

=2\,tan^2A+2+5\,tanA=2tan2A+2+5tanA

=2(tan^2A+1)+5\,tanA=2(tan2A+1)+5tanA

Using,

{\bf\,sec^2\theta-tan^2\theta=1\implies\,sec^2\theta=1+tan^2\theta}sec2θ−tan2θ=1⟹sec2θ=1+tan2θ

we get

=2\,sec^2A+5\,tanA=2sec2A+5tanA+tanA+2)(2tanA+1)=2\,sec^2A+5\,tanA}⟹(tanA+2)(2tanA+1)=2sec2A+5tanA