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# Our school exam need substitution methodlike, x = sin1/2θ, cosΘ

147 Points
11 years ago

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

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