(tanθ+2)(2tanθ+1)=5tanθ+sec2θ. Write ‘True’ or ‘False’ and justify your answer.
Trigonometric identity:
L.H.S.(tanθ+2)(2tanθ+1)
=2tan2θ+tanθ+4tanθ+2
=2tan2θ+5tanθ+2
=2sec2θ-1+5tanθ+2[∵1+tan2θ=sec2θ]
=2sec2θ-2+5tanθ+2 [∵1+tan2θ=sec2θ] =2sec2θ+5tanθ
≠R.H.S
∴L.H.S≠R.H.S
Hence, the given statement is false.