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प्रश्न
Find the general solution of : sinx · tanx = tanx - sinx + 1
बेरीज
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उत्तर
sinx · tanx = tanx - sinx + 1
`sinx.(sinx/cosx) - 1 = sin x/cos x - sin x`
`[ sin^2x - cos x ]/cos x = [ sin x - sin x. cos x]/cos x`
`sin^2x - cos x = sin x - sin x.cosx`
now x can't equal to `pi/2`
`sin^2x - cosx - sin x + sinxcosx = 0`
`sin^2x + sinxcosx - cosx - sinx = 0`
`sinx( sinx + cos x) - 1(sinx + cosx ) = 0`
`(sinx - 1)(sinx + cosx) = 0`
since x is not `pi/2`
sin x + cosx = 0
`sin x = - cos x`
`x = "multiple of" pi/4 "in even quads."`
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