if fn(θ)=tanθ2(1+sec⁡θ)(1+sec⁡2θ)…1+sec⁡2nθ,then

if fn(θ)=tanθ2(1+secθ)(1+sec2θ)1+sec2nθ,then

  1. A

    f2π16=1

  2. B

    f3π32=1

  3. C

    f4π64=1

  4. D

    f5π128=1

    Register to Get Free Mock Test and Study Material

    +91

    Verify OTP Code (required)

    I agree to the terms and conditions and privacy policy.

    Solution:

    f0(θ)=tanθ2(1+secθ)

    =sinθ2cosθ21+1cosθ=sinθ2cosθ2(cosθ+1)cosθ

    =2sinθ2cosθ2cosθ=sinθcosθ=tanθf1(θ)=tanθ(1+sec2θ)=sinθcosθ(cos2θ+1)cos2θ=tan2θLikewisefn(θ)=tan2nθf2π16=tan22π16=1,f3π32=tan23π32=1f4π64=tan24π64=1,f5π128=tan25π128=1

    Chat on WhatsApp Call Infinity Learn

      Register to Get Free Mock Test and Study Material

      +91

      Verify OTP Code (required)

      I agree to the terms and conditions and privacy policy.