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(cosec^2*theta)*(sin(theta)^4 - cos(theta)^4 %2B 1)=2 |
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Answer» (sin⁴θ-cos⁴θ+1)cosec²θ=[{(sin²θ)²-(cos²θ)²}+1]cosec²θ=[{(sin²θ+cos²θ)(sin²θ-cos²θ)}+1]cosec²θ=(sin²θ-cos²θ+1)cosec²θ [Using sin²θ+cos²θ=1]={sin²θ+(1-cos²θ)}cosec²θ=(sin²θ+sin²θ)cosec²θ=2sin²θ.cosec²θ=2sin²θ×1/sin²θ=2 (Proved) (sin4θ - cos4θ + 1) cosec²θ =[ (sin²θ+cos²θ) (sin²θ-cos²θ) + 1] cosec²θ =[1 (sin²θ - cos²θ) +1] cosec²θ =[ sin²θ - (1-sin²θ) +1] cosec²θ =(sin²θ -1 +sin²θ +1 )cosec²θ =2sin²θ cosec²θ=2 =R.H.S HENCE PROVED |
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