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Prove that `tan^-1((cosx)/(1+sinx))=pi/4-x/2, x in (-pi/2,pi/2)` |
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Answer» Here, we will use the folowing formulas: `cos2x = cos^2x-sin^2x` `sin2x = 2sinxcosx` `tan(pi/4-x) = (1-tanx)/(1+tanx)` Now, `L.H.S. = tan^-1(cosx/(1+sinx))` `= tan^-1((cos^2(x/2)-sin^2(x/2))/(cos^2(x/2)+sin^2(x/2)+2sin(x/2)cos(x/2)))` `=tan^-1(((cos(x/2)+sin(x/2))(cos(x/2)-sin(x/2)))/(cos(x/2)+sin(x/2))^2)` `=tan^-1((cos(x/2)-sin(x/2))/(cos(x/2)+sin(x/2)))` `=tan^-1((1-sin(x/2)/(cos(x/2)))/(1+sin(x/2)/cos(x/2)))` `=tan^-1((1-tan(x/2))/(1+tan(x/2)))` `=tan^-1(tan(pi/4-x/2))` `=(pi/4-x/2) = R.H.S.` |
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