If cos(θ−α)=a,sin(θ−β)=b, where (θ−α,θ−β)∈(0,π2) then-
A
sin(α−β)=ab−√(1−a2−b2+a2b2)
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B
cos(α−β)=a√(1−b2)+b√(1−a2)
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C
sin(α−β)=ab+√(1−a2−b2+a2b2)
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D
cos(α−β)=a√(1−b2)−b√(1−a2)
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Solution
The correct options are Asin(α−β)=ab−√(1−a2−b2+a2b2) Bcos(α−β)=a√(1−b2)+b√(1−a2) Given cos(θ−α)=a,sin(θ−β)=b, where (θ−α,θ−β)ϵ(0,π2) Now, sin(α−β)=sin((θ−β)−(θ−α)) =sin(θ−β)cos(θ−α)−cos(θ−β)sin(θ−α) =ab−√1−a2√1−b2 ⇒sin(α−β)=ab−√(1−a2−b2+a2b2) cos(α−β)=cos((θ−β)−(θ−α)) =cos(θ−β)cos(θ−α)+sin(θ−β)sin(θ−α) =a√1−b2+b√1−a2 cos(α−β)=a√(1−b2)+b√(1−a2)