1/6sinx ,cosx ,tanx are in gp then the value of x
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1/sinx , cosx , tanx are in GP
so,
cosx/(1/sinx) = tanx/(cosx)
sinx.cosx = sinx/cos²x
sinx( cos³x -1) = 0
but sinx ≠ 0 becoz in this way 1/sinx =∞
so, cos³x= 1
cosx = 1
x = 2nπ where n is integer number
so,
cosx/(1/sinx) = tanx/(cosx)
sinx.cosx = sinx/cos²x
sinx( cos³x -1) = 0
but sinx ≠ 0 becoz in this way 1/sinx =∞
so, cos³x= 1
cosx = 1
x = 2nπ where n is integer number
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