Question

Prove that (1 + sec 2θ)(1 + sec 4θ).............(1 + sec 2nθ) = tan 2n θ cot θ

Answer 1

we have to prove that,

(1 + sec 2θ)(1 + sec 4θ).............(1 + sec 2ⁿθ) = tan 2ⁿθ.cotθ

or, tanθ(1 + sec 2θ)(1 + sec 4θ).............(1 + sec 2ⁿθ) = tan 2ⁿθ

now, f(θ) = tanθ.(1 + sec2θ)

= (sinθ/cosθ)(1 + cos2θ)/cos2θ

= (sinθ/cosθ) (2cos²θ)/cos2θ

= (2sinθ.cosθ)/cos2θ

= sin2θ/cos2θ

= tan2¹θ

similarly, f(2θ) = tan2θ.(1 + sec4θ)

= (sin2θ/cos2θ) (1 + cos4θ)/cos4θ

= (sin2θ/cos2θ)(2cos²2θ)/cos4θ

= (2sin2θ.cos2θ)/cos4θ

= tan4θ = tan2²θ

f(3θ) = tan4θ(1 + sec8θ) = tan8θ = tan2³θ

f(4θ) = tan8θ(1 + sec16θ) = tan16θ = tan2⁴θ

....... ..... ........

f(nθ) =tan2ⁿ-¹θ(1 + sec2ⁿθ) = tan2ⁿθ

⇒ f(nθ) = tanθ(1 + sec 2θ)(1 + sec 4θ).............(1 + sec 2ⁿθ) = tan 2ⁿθ

hence, it is clear that, tanθ(1 + sec 2θ)(1 + sec 4θ).............(1 + sec 2ⁿθ) = tan 2ⁿθ

or, (1 + sec 2θ)(1 + sec 4θ).............(1 + sec 2ⁿθ) = tan 2ⁿθ.cotθ