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Question Number 182848 by malithxd last updated on 15/Dec/22

((sin^3 x−sin x+cos x)/(sin x))=cot x −cos^2 x  prove this

$$\frac{{sin}^{\mathrm{3}} {x}−{sin}\:{x}+{cos}\:{x}}{{sin}\:{x}}={cot}\:{x}\:−{cos}^{\mathrm{2}} {x} \\ $$$${prove}\:{this} \\ $$$$ \\ $$$$ \\ $$

Answered by Rasheed.Sindhi last updated on 15/Dec/22

((sin^3 x−sin x+cos x)/(sin x))=cot x −cos^2 x  LHS:sin^2 x−1+((cos x )/(sin x))                =−(1−sin^2 x)+((cos x )/(sin x))               − cos^2 x+cot x=RHS

$$\frac{{sin}^{\mathrm{3}} {x}−{sin}\:{x}+{cos}\:{x}}{{sin}\:{x}}={cot}\:{x}\:−{cos}^{\mathrm{2}} {x} \\ $$$${LHS}:\mathrm{sin}^{\mathrm{2}} {x}−\mathrm{1}+\frac{\mathrm{cos}\:{x}\:}{\mathrm{sin}\:{x}}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=−\left(\mathrm{1}−\mathrm{sin}^{\mathrm{2}} {x}\right)+\frac{\mathrm{cos}\:{x}\:}{\mathrm{sin}\:{x}}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:−\:\mathrm{cos}^{\mathrm{2}} {x}+\mathrm{cot}\:{x}={RHS}\: \\ $$

Answered by manxsol last updated on 15/Dec/22

/sinx  sin^2 −1+ctx  −cos^2 +cotx=cotx−cos^2 x

$$/{sinx} \\ $$$${sin}^{\mathrm{2}} −\mathrm{1}+\boldsymbol{{ctx}} \\ $$$$−\boldsymbol{{cos}}^{\mathrm{2}} +\boldsymbol{{cotx}}=\boldsymbol{{cotx}}−\boldsymbol{{cos}}^{\mathrm{2}} \boldsymbol{{x}} \\ $$$$ \\ $$

Answered by cortano1 last updated on 15/Dec/22

 ((sin^3 x−sin x+cos x)/(sin x))   = ((sin x(sin^2 x−1)+cos x)/(sin x))  = sin^2 x−1 + cot x  =−cos^2 x + cot x

$$\:\frac{\mathrm{sin}\:^{\mathrm{3}} {x}−\mathrm{sin}\:{x}+\mathrm{cos}\:{x}}{\mathrm{sin}\:{x}}\: \\ $$$$=\:\frac{\mathrm{sin}\:{x}\left(\mathrm{sin}\:^{\mathrm{2}} {x}−\mathrm{1}\right)+\mathrm{cos}\:{x}}{\mathrm{sin}\:{x}} \\ $$$$=\:\mathrm{sin}\:^{\mathrm{2}} {x}−\mathrm{1}\:+\:\mathrm{cot}\:{x} \\ $$$$=−\mathrm{cos}\:^{\mathrm{2}} {x}\:+\:\mathrm{cot}\:{x}\: \\ $$

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