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Prove that \frac{\sin \theta-\cos \theta+1}{\sin \theta+\cos \theta-1}=\frac{1}{\sec \theta-\tan \theta} using the identity \sec ^{2} \theta=1+\tan ^{2} \theta |
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Answer» sinx-cosx+1/ sinx+cosx -1 =(sinx-cosx+1)x(sinx +cosx +1) / (sinx+cosx - 1)x(sinx +cosx +1) =(sinx +1)sq. - (cosx)sq./ (sinx +cosx) sq. - (1)sq. =sinsq.x+2sinx+1-cossq.x/ sinsq.x+cossq.x+2sinxcosx-1 =2sinsq.x+2sinx/ 2sinxcosx =2sinx(sinx +1)/ 2sinxcosx =sinx+1/cosx =(1+sinx)x(1-sinx)/ (cosx) x(1-sinx) =cosx/(1-sinx) (dividing num. and den. by cosx) =1/secx-tanx = sec x + tan x |
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