Question

Prove that (tanθ + 2) (2 tanθ + 1) = 5 tanθ + sec2 θ + 2​

Answer 1

Step-by-step explanation:

I believe your Question was,

"Prove that

(tanθ + 2)(2 tanθ + 1) = 5 tanθ + 2sec²θ."

LHS = (tanθ + 2)(2 tanθ + 1)

= 2tan²θ + tanθ + 4tanθ + 2

= 2tan²θ + 5tanθ + 2

We know that,

1 + tan²θ = sec²θ

Thus,

tan²θ = sec²θ - 1

Putting that in equation we get,

= 2(sec²θ - 1) + 5tanθ + 2

= 2sec²θ - 2 + 5tanθ + 2

= 5tanθ + 2sec²θ + 2 - 2

= 5 tanθ + 2sec²θ = RHS

Thus,

LHS = RHS

Hence proved.

Hope it helped and you understood it........All the best