( \operatorname { tan } \theta + 2 ) ( 2 \operatorname { tan } \theta

1.	( \operatorname { tan } \theta + 2 ) ( 2 \operatorname { tan } \theta + 1 ) = 5 \operatorname { tan } \theta + 2 \operatorname { sec } ^ { 2 } \theta
Answer» L.H.S., (tanθ + 2)(2tanθ + 1) = 2tan^2θ + tanθ + 4tanθ + 2 = 5tanθ + 2tan^2θ + 2 = 5tanθ + 2 (tan^2θ + 1) = 5tanθ + 2 (sec^2θ) [Since, tan^2θ + 1 = sec^2θ] = 5tanθ + 2sec^2θ = R.H.S.

( \operatorname { tan } \theta + 2 ) ( 2 \operatorname { tan } \theta + 1 ) = 5 \operatorname { tan } \theta + 2 \operatorname { sec } ^ { 2 } \theta

Answer»

L.H.S., (tanθ + 2)(2tanθ + 1)

= 2tan^2θ + tanθ + 4tanθ + 2 = 5tanθ + 2tan^2θ + 2 = 5tanθ + 2 (tan^2θ + 1) = 5tanθ + 2 (sec^2θ)

[Since, tan^2θ + 1 = sec^2θ] = 5tanθ + 2sec^2θ = R.H.S.