How do you find a numerical value of one trigonometric function of x given $cotx+sinx=-cosxcotx$ ?

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Answer 1 · 2016-11-12T10:29:07

$:.x=+-pi+2pin$

$cotx+sinx=-cosxcotx$

$cos x/sin x+sinx=-cosx cosx/sinx$

$(cos x+sin^2x)/sinx=-cos^2x/sinx$

$(cos x+sin^2x)/sinx *sinx=-cos^2x/sinx *sinx$

$(cos x+sin^2x)/cancelsinx *cancelsinx=-cos^2x/cancelsinx *cancelsinx$

$cos x+sin^2x=-cos^2x$

$cos x+sin^2x+cos^2x=0$

$cosx+1=0$

$cosx=-1$

$x=cos^-1 (-1)$

$:.x=+-pi+2pin$

How do you find a numerical value of one trigonometric function of x given cotx+sinx=-cosxcotxcotx+sinx=-cosxcotx?