If cos⁡(θ−α)=a and  cos⁡(θ−β)=b then sin2⁡(α−β)+2abcos⁡(α−β) is equal to

If cos⁡(θ−α)=a and  cos(θβ)=b then sin2(αβ)+2abcos(αβ) is equal to

  1. A

    a2+b2

  2. B

    a2-b2

  3. C

    b2a2

  4. D

    -a2-b2

    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:

    αβ=(θβ)(θα)cos(αβ)=cos((θβ)(θα))=cos(θβ)cos(θα)+sin(θβ)sin(θα) and  sin(αβ)=sin(θβ)cos(θα)+cos(θβ)sin(θα) cos(αβ)=ba+1a21b2 .................(1) and  sin(αβ)=a1b2b1a2 Now,  sin2(αβ)=a1b2+b1a22 sin2(αβ)=a21b2+b21a22ab{cos(αβ)ab}           from eq (1) sin2(αβ)+2abcos(αβ)=a2a2b2+b2b2a2+2a2b2 sin2(αβ)+2abcos(αβ)=a2+b2

    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.