if y=log tan(π/4+x/2), show that dy/dx=secx
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heya......
dy/dx={1/tan (pi/4+x/2)}sec^2 (pi/4+x/2).1/2
=1/2sec(pi/4+x/2)/sin (pi/4+x/2)
=1/{2sin (pi/4+x/2).cos (pi/4+x/2)}
=1/sin (pi/2+x)=1/cosx=secx
hence
dy/dx=secx
dy/dx-secx=0
tysm.@gozmit
dy/dx={1/tan (pi/4+x/2)}sec^2 (pi/4+x/2).1/2
=1/2sec(pi/4+x/2)/sin (pi/4+x/2)
=1/{2sin (pi/4+x/2).cos (pi/4+x/2)}
=1/sin (pi/2+x)=1/cosx=secx
hence
dy/dx=secx
dy/dx-secx=0
tysm.@gozmit
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Explanation:
A ladder rest against a wall at an angle α to the the horizontal.Its foot is pulled away from the wall through a distance a,so that it slides a distance b down the wall making an angle with the horizontal.
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