InterviewSolution
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InterviewSolution
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Differentiate w.r.t. x: `(i)tan^(-1)((cosx)/(1+sinx))" "(ii)tan^(-1)(secx+tanx)` |
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Answer» Let `y=tan^(-1)((cos+1)/(1+sinx))=tan^(-1){(sin((pi)/(2)-x))/(1+cos((pi)/(2)-x))}` `=tan^(-1){(2sin((pi)/(4)-(x)/(2))cos((pi)/(4)-(x)/(2)))/(2cos^(2)((pi)/(4)-(x)/(2))}}` `tan^(-1){tan((pi)/(4)-(x)/(2))}=((pi)/(4)-(x)/(2)).` `thereforey=((pi)/(4)-(x)/(2)).` Hence,`(dy)/(dx)=(d)/(dx)((pi)/(4)-(x)/(2))=(d)/(dx)((pi)/(4))-(d)/(dx)((x)/(2))=(0-(1)/(2))=(-1)/(2).` (ii) Let `y=tan^(-1)(sec x+tanx)` `=tan^(-1)((1)/(cosx)+(sinx)/(cosx))=tan^(-1)((1+sinx)/(cosx))` `=tan^(-1){(1-cos((pi)/(2)+x))/(sin((pi)/(2)+x))}` `{because cos ((pi)/(2)+x)=-sinx,sin((pi)/(2)+x)=cosx}` `=tan^(-1){(2sin^(2)((pi)/(4)+(x)/(2)))/(2sin((pi)/(4)+(x)/(2))cos((pi)/(4)+(x)/(2)))}` `=tan^(-1){tan((pi)/(4)+(x)/(2))}=((pi)/(4)+(x)/(2)).` `therefore y=((pi)/(4)+(x)/(2)).` Hence, `(dy)/(dx)=(d)/(dx)((pi)/(4)+(x)/(2))=(d)/(dx)((pi)/(4))+(d)/(dx)((x)/(2))=(0+(1)/(2))=(1)/(2).` |
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