Question

sinx tanx -1=tanx-sinx

Answer 1

$sinx \: tanx \: - 1 = tanx - sinx \\ {sin}^{2} x( \frac{1}{cosx} ) - 1 = \frac{sinx}{cosx} - sinx \\ \frac{ {sin}^{2} x - cosx}{cosx} = \frac{sinx - sinx \: cosx}{cosx } \\ {sin}^{2} x - cosx = sinx - sinxcosx \\ now \: x \: cant \: \: equal \: to \: \frac{\pi}{2} \\ {sin}^{2} x - cosx - sinx + sinx \: cosx = 0 \\ {sin}^{2} x + sinxcosx - cosx - sinx = 0 \\ sinx(sinx + cosx) - 1(sinx + cosx) = 0 \\ (sinx - 1)(sinx + cosx) = 0 \\ since \: x \: is \: not \: \frac{\pi}{2} \\ sinx + cosx = 0 \\ sinx = - cosx \\ x =multiple \: of \: \frac{\pi}{4} in \: even \: quads$

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Answer 2

Sinx.tanx-1=tanx-sinx

-1=tanx-sinx-sinx.tanx

Sinx-1=tanx-sinx.tanx

Sinx-1=tanx(1-sinx)

-(1-sinx)=tanx(1-sinx)

-1=tanx

WE KNOW THAT

tanx=tany

x=nπ+y

SO THAT

tanx=tanπ/4

x=nπ+y

Solved

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