If sin⁡ (θ+α)=a,cos2 ⁡(θ+β)=b, then sin ⁡(α−β)=

If sin (θ+α)=a,cos2 (θ+β)=b, then sin (αβ)=

  1. A

    ab1a21b2

  2. B

    ab1a21b2

  3. C

    ±ab1a2(1b)

  4. D

    ba(1a)2(1b)

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    Solution:

    We have,

    sin(θ+α)=a and cos(θ+β)=±b cos(θ+α)=±1a2 and, sin(θ+β)=±1b

    Now, 

    sin(αβ)=sin{(θ+α)(θ+β)}sin(αβ)=sin(θ+α)cos(θ+β)cos(θ+α)sin(θ+β) sin(αβ)=±ab±1a2×{±1b} sin(αβ)=±ab1a21b

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