Question

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Answer 1

On canceling (+sin2αsin2β) and (−sin2αsin2β),

:⟹2sin

2

β−(2sin

2

α)(2sin

2

β)+cos2αcos2β

We know that,

:↪2sin

2

θ=1−cos2θ

So,

:⟹(2sin

2

β)−(2sin

2

α)(2sin

2

β)+cos2αcos2β

:⟹(1−cos2β)−(1−cos2α)(1−cos2β)+cos2αcos2β

:⟹1−cos2β−(1−cos2β−cos2α+cos2αcos2β)+cos2αcos2β