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tan(90+x)=tan(x-50)tan(x)tan(x+50) wht is the min value of ''x''?
tan (90+x)/tan x = tan( x+50). tan (x-50)/1 using componendo and dividendo priciple , we have (tan (90+x) + tan x) /( tan (90+x) - tan x )= (tan( x+50). tan (x-50) + 1) / (tan( x+50). tan (x-50) - 1) Sin (90+2x) / sin 90 = cos 100 / cos 2x here sin 90 = 1 so, by multiplying 2 on both sides we have 2 sin(2x+90).cos 2x + 2 cos 100 = 0 sin (4x+100) + 1+ 2cos 100 =0 on simplification gives cos 4x + 1 - 2.sin 10 = 0 2* cos22x = 2* sin 10 so, cos22x = sin 10 so , x = cos-1[(sin 10)1/2] / 2
tan (90+x)/tan x = tan( x+50). tan (x-50)/1
using componendo and dividendo priciple , we have
(tan (90+x) + tan x) /( tan (90+x) - tan x )= (tan( x+50). tan (x-50) + 1) / (tan( x+50). tan (x-50) - 1)
Sin (90+2x) / sin 90 = cos 100 / cos 2x
here sin 90 = 1
so, by multiplying 2 on both sides we have
2 sin(2x+90).cos 2x + 2 cos 100 = 0
sin (4x+100) + 1+ 2cos 100 =0
on simplification gives
cos 4x + 1 - 2.sin 10 = 0
2* cos22x = 2* sin 10
so, cos22x = sin 10
so , x = cos-1[(sin 10)1/2] / 2
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