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प्रश्न
If y = log `[tan(pi/4+x/2)]`Prove that
I. tan h`y/2 = tan pi/2`
II. cos hy cos x = 1
उत्तर
1] `sin hy/2=(e^(y/2)-e^(-y/2))/2`
`cos h y/2=(e^(y/2)+e^((-y)/2))/2`
`tan h y/2=(sin hy/2)/(cos h y/2)`
`=((e^(y/2)-e^((-y)/2))/2)/((e^(y/2)-e^(-y/2))/2)`
`= (e^y-1)/(e^y+1)`
But `e^U = (1+tan x/2)/(1-tan x/2)`
`therefore tan hy/2=( (1+tan x/2)/(1-tan x/2)-1)/((1+tan x/2)/(1-tan x/2)+1)`
`=(1+tan x/2-1+tan x/2)/(1+tan x/2+1-tan x/2)`
`=tan x/2`
`therefore tan hy/2=tan x/2`
2] `y=log[tan (pi/4+x/2)]`
`e^y=tan(pi/4+x/2)`
`=(tan pi/4+tan x/2)/(1-tan pi/4tan pi/2)`
`e^y=(1+tan x/2)/(1-tan x/2)`
`e^(-y)=(1-tan x/2)/(1+tan x/2)`
cos hy = `(e^y+e^(-y))/2`
`=((1+tan x/2)/(1-tan x/2)+(1-tan x/2)/(1+tan x/2))/2`
`= ((1+tan x/2)^2+(1-tan x/2)^2)/(2(1-tan^2 x/2))`
`=(1+2tan x/2+tan^2 x/2+1+tan^2 x/2-2tan x/2)/(2(1-tan^2 x/2))`
`= (1+tan^2 x/2)/(1-tan^2 x/2)`
`therefore cos hy=1/(cosx)`
`therefore cos hy cosx=1/(cosx).cosx`
cos hy cos x = 1
Hence proved
