For a positive integer n , let fn(θ) =[tanθ/2](1+secθ)(1+sec2θ)(1+sec4θ)...(1+sec2nθ) Then f2 [pi/16]=1 f3[pi/32]=1 f4[pi/64]=1 f5[pi/128]=1

by induction we get fn(theta)=tan(2ntheta)so all are correct

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Trigonometry> For a positive integer n , let f n (θ) =[...

For a positive integer n , let f_n(θ) =[tanθ/2](1+secθ)(1+sec2θ)(1+sec4θ)...(1+sec2ⁿθ) Then

f₂ [pi/16]=1

f₃[pi/32]=1

f₄[pi/64]=1

f₅[pi/128]=1

Ruchi , 11 Years ago

Grade 11

2 Answers

rajesh

by induction we get f_n(theta)=tan(2ⁿtheta)

so all are correct

Last Activity: 11 Years ago

Nilesh Kumar

tanθ/2(1+1/cosθ)(1+1/cos2θ)...tanθ/2(cosθ+1)(1/cosθ)(cos2θ+1)(1/cos2θ).tanθ/2(2cos²θ/2)(1/cosθ)(2cos²θ)(1/cos2θ)=tan2θSo we can solving by taking(1+sec4θ) we will get tan4θ. So we can conclude that fn(theta)=tan(2^ntheta) so all the options are satisfying it. CORRECT ANSWER-A,B,C and D

Last Activity: 8 Years ago