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प्रश्न
Find the value of the following:
`tan^(-1)(1) + cos^(-1) (-1/2) + sin^(-1) (-1/2)`
उत्तर १
Let `tan^(-1) (1)` = x Then tan x= 1 = tan `pi/4`
`:. tan^(-1) (1) = pi/4`
Let `cos^(-1) (-1/2) = y` Then, `cos y = -1/2 = -cos(pi/3) = cos(pi - pi/3) = cos ((2pi)/3)`
`:. cos^(-1) (- 1/2) = (2pi)/3`
Let `sin^(-1) (-1/2) = z`. Then `sin z =-1/2 = -sin(pi/6) = sin(-pi/6)`
`:. sin^(-1)(-1/2) = - pi/6`
`:. tan^(-1) (1) + cos^(-1) (-1/2) + sin^(-1) (-1/2)`
`= pi/4 + (2pi)/3 - pi/6`
`= (3pi + 8pi - 2pi)/12 `
`= (9pi)/12 = (3pi)/4`
उत्तर २
`"tan"^-1 (1) + "cos"^-1 (-1/2) + "sin"^-1 (-1/2)`
`= "tan"^-1 ("tan" pi/4) + "cos"^-1 ("cos" (2pi)/3) + "sin"^-1 "sin" ((-pi)/6)`
`= pi/4 + (2pi)/3 + ((-pi)/6)`
`= (3pi + 8pi - 2pi)/12`
`= (9 pi)/12 = (3 pi)/4`
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