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Question Number 75173 by liki last updated on 08/Dec/19

Commented by liki last updated on 08/Dec/19

...i need help plz

$$...{i}\:{need}\:{help}\:{plz} \\ $$

Answered by mind is power last updated on 08/Dec/19

sin(θ)=θ+o(θ^2 )  3tan(θ)=3θ+o(θ^2 )  cos(θ)=1−(θ^2 /2)+o(θ^2 )  5+3tan(θ)−4cos(θ)=1+3θ+2θ^2 +o(θ^2 )  ((θ+o(θ^2 ))/(1+(2θ^2 +3θ+o(θ))))=θ(1−(2θ^2 +3θ+o(θ))  =θ−3θ^2 +o(θ^2 )

$$\mathrm{sin}\left(\theta\right)=\theta+\mathrm{o}\left(\theta^{\mathrm{2}} \right) \\ $$$$\mathrm{3tan}\left(\theta\right)=\mathrm{3}\theta+\mathrm{o}\left(\theta^{\mathrm{2}} \right) \\ $$$$\mathrm{cos}\left(\theta\right)=\mathrm{1}−\frac{\theta^{\mathrm{2}} }{\mathrm{2}}+\mathrm{o}\left(\theta^{\mathrm{2}} \right) \\ $$$$\mathrm{5}+\mathrm{3tan}\left(\theta\right)−\mathrm{4cos}\left(\theta\right)=\mathrm{1}+\mathrm{3}\theta+\mathrm{2}\theta^{\mathrm{2}} +\mathrm{o}\left(\theta^{\mathrm{2}} \right) \\ $$$$\frac{\theta+\mathrm{o}\left(\theta^{\mathrm{2}} \right)}{\mathrm{1}+\left(\mathrm{2}\theta^{\mathrm{2}} +\mathrm{3}\theta+\mathrm{o}\left(\theta\right)\right)}=\theta\left(\mathrm{1}−\left(\mathrm{2}\theta^{\mathrm{2}} +\mathrm{3}\theta+\mathrm{o}\left(\theta\right)\right)\right. \\ $$$$=\theta−\mathrm{3}\theta^{\mathrm{2}} +\mathrm{o}\left(\theta^{\mathrm{2}} \right) \\ $$

Commented by liki last updated on 08/Dec/19

...thank you sir.

$$...{thank}\:{you}\:{sir}. \\ $$

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