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Question Number 93397 by Rio Michael last updated on 12/May/20
prove that ((1−tan^3 θ)/(1 + tan^3 θ)) = 1−2sin^3 θ
$$\mathrm{prove}\:\mathrm{that}\:\frac{\mathrm{1}−\mathrm{tan}^{\mathrm{3}} \theta}{\mathrm{1}\:+\:\mathrm{tan}^{\mathrm{3}} \theta}\:=\:\mathrm{1}−\mathrm{2sin}^{\mathrm{3}} \theta\: \\ $$
Commented by john santu last updated on 13/May/20
θ = 45^(o ) ⇒((1−1)/(1+1)) = 0 [ lhs]   θ=45^o ⇒1−2(((√2)/2))^3 = 1−2(((2(√2))/8))≠0 [rhs]  the question wrong
$$\theta\:=\:\mathrm{45}^{\mathrm{o}\:} \Rightarrow\frac{\mathrm{1}−\mathrm{1}}{\mathrm{1}+\mathrm{1}}\:=\:\mathrm{0}\:\left[\:\mathrm{lhs}\right]\: \\ $$$$\theta=\mathrm{45}^{\mathrm{o}} \Rightarrow\mathrm{1}−\mathrm{2}\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)^{\mathrm{3}} =\:\mathrm{1}−\mathrm{2}\left(\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{8}}\right)\neq\mathrm{0}\:\left[\mathrm{rhs}\right] \\ $$$$\mathrm{the}\:\mathrm{question}\:\mathrm{wrong}\: \\ $$$$ \\ $$
Commented by Rio Michael last updated on 13/May/20
thank you sir
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

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