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Our school exam need substitution method
like, x = sin1/2θ, cosΘ
Dear aby you can also solve by subtituting x2=sin2Θ tan-1[(1+x2)1/2 - (1-x2)1/2]/[(1+x2)1/2 + (1-x2)1/2] =tan-1[(1+sin2Θ)1/2 - (1-sin2Θ)1/2]/[(1+sin2Θ)1/2 + (1-sin2Θ)1/2] now 1+sin2Θ =sin2Θ +cos2Θ + 2sinΘ cosΘ 1+sin2Θ =(sinΘ + cosΘ)2 similerly 1-sin2Θ =(sinΘ - cosΘ)2 put these value tan-1[(sinΘ+cosΘ) - (sinΘ -cosΘ)]/[(sinΘ+cosΘ) + (sinΘ -cosΘ)] or = tan-1 cotΘ = ∏/2 - Θ =∏/2 - 1/2 sin-1 x2 now differentiate and you will get the result Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE & AIEEE preparation. All the best. Regards,Askiitians ExpertsBadiuddin
Dear aby
you can also solve by subtituting x2=sin2Θ
tan-1[(1+x2)1/2 - (1-x2)1/2]/[(1+x2)1/2 + (1-x2)1/2]
=tan-1[(1+sin2Θ)1/2 - (1-sin2Θ)1/2]/[(1+sin2Θ)1/2 + (1-sin2Θ)1/2]
now 1+sin2Θ =sin2Θ +cos2Θ + 2sinΘ cosΘ
1+sin2Θ =(sinΘ + cosΘ)2
similerly 1-sin2Θ =(sinΘ - cosΘ)2
put these value
tan-1[(sinΘ+cosΘ) - (sinΘ -cosΘ)]/[(sinΘ+cosΘ) + (sinΘ -cosΘ)]
or = tan-1 cotΘ
= ∏/2 - Θ
=∏/2 - 1/2 sin-1 x2
now differentiate and you will get the result
Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the
answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE & AIEEE preparation.
All the best. Regards,Askiitians ExpertsBadiuddin
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