y=tan^-1(cosx/1+sinx)
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hii friend ....
here is your answer♥♥♥♥♥
y=tan^-1 (cosx/1+sinx)
=tan^-1 cos^2 x/2-sin^2 x/2 / (cosx/2+sinx/2)
=tan^-1 [cosx/2-sinx/2/cosx/2+sinx/2]
=tan^-1 [1-tanx/2/1+tanx/2]
=tan^-1 [tan ¥/4-tan x/2/1+tan ¥/4.tanx/2]
=tan^-1 tan (¥/4-x/2)
=¥/4-x/2
dy/dx= -1/2
¥ means pai
here is your answer♥♥♥♥♥
y=tan^-1 (cosx/1+sinx)
=tan^-1 cos^2 x/2-sin^2 x/2 / (cosx/2+sinx/2)
=tan^-1 [cosx/2-sinx/2/cosx/2+sinx/2]
=tan^-1 [1-tanx/2/1+tanx/2]
=tan^-1 [tan ¥/4-tan x/2/1+tan ¥/4.tanx/2]
=tan^-1 tan (¥/4-x/2)
=¥/4-x/2
dy/dx= -1/2
¥ means pai
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