Question Number 93397 by Rio Michael last updated on 12/May/20
$$\mathrm{prove}\:\mathrm{that}\:\frac{\mathrm{1}−\mathrm{tan}^{\mathrm{3}} \theta}{\mathrm{1}\:+\:\mathrm{tan}^{\mathrm{3}} \theta}\:=\:\mathrm{1}−\mathrm{2sin}^{\mathrm{3}} \theta\: \\ $$
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](https://www.tinkutara.com/question/Q93437.png)
$$\theta\:=\:\mathrm{45}^{\mathrm{o}\:} \Rightarrow\frac{\mathrm{1}−\mathrm{1}}{\mathrm{1}+\mathrm{1}}\:=\:\mathrm{0}\:\left[\:\mathrm{lhs}\right]\: \\ $$$$\theta=\mathrm{45}^{\mathrm{o}} \Rightarrow\mathrm{1}−\mathrm{2}\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)^{\mathrm{3}} =\:\mathrm{1}−\mathrm{2}\left(\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{8}}\right)\neq\mathrm{0}\:\left[\mathrm{rhs}\right] \\ $$$$\mathrm{the}\:\mathrm{question}\:\mathrm{wrong}\: \\ $$$$ \\ $$
Commented by Rio Michael last updated on 13/May/20
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$