For a positive integer n , let f n (θ) =[tanθ/2](1+secθ)(1+sec2θ)(1+sec4θ)...(1+sec2 n θ) Then f 2 [pi/16]=1 f 3 [pi/32]=1 f 4 [pi/64]=1 f 5 [pi/128]=1

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


2 Answers

30 Points
7 years ago
by induction we get fn(theta)=tan(2ntheta)
so all are correct
Nilesh Kumar
13 Points
5 years ago
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

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