Question

question 32 pls help me.....​

Answer 1

Answer:

(tanθ+2)(2tanθ+1)=5tanθ+2sec²θ

LHS=(tanθ+2)(2tanθ+1)

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

2tan²θ+tanθ+4tanθ+2

2tan²θ+5tanθ+2

2(sec²θ-1)+5tanθ+2

2sec²θ-2+5tanθ+2

5tanθ+2sec²θ-2+2

5tanθ+2sec²θ=RHS

Answer 2

Answer:

Step-by-step explanation:

FOLLOW ME ON INSTAGRAM

@muhammed10leo