Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 120468 by help last updated on 31/Oct/20

Commented by bemath last updated on 01/Nov/20

cot (θ+30°) cot (θ−30°)=  (1/(tan (θ+30°) tan (θ−30°))) =  (1/((((tan θ+((√3)/3))/(1−(((√3) tan θ)/3))) . ((tan θ−((√3)/3))/(1+(((√3) tan θ)/3)))))) =  ((1−(1/3)tan^2 θ)/(tan^2 θ−(1/3))) = ((3−tan^2 θ)/(3tan^2 θ−1)) : [ ((tan^2 θ)/(tan^2 θ)) ]  = ((3cot^2 θ−1)/(3−cot^2 θ))

$$\mathrm{cot}\:\left(\theta+\mathrm{30}°\right)\:\mathrm{cot}\:\left(\theta−\mathrm{30}°\right)= \\ $$$$\frac{\mathrm{1}}{\mathrm{tan}\:\left(\theta+\mathrm{30}°\right)\:\mathrm{tan}\:\left(\theta−\mathrm{30}°\right)}\:= \\ $$$$\frac{\mathrm{1}}{\left(\frac{\mathrm{tan}\:\theta+\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}}{\mathrm{1}−\frac{\sqrt{\mathrm{3}}\:\mathrm{tan}\:\theta}{\mathrm{3}}}\:.\:\frac{\mathrm{tan}\:\theta−\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}}{\mathrm{1}+\frac{\sqrt{\mathrm{3}}\:\mathrm{tan}\:\theta}{\mathrm{3}}}\right)}\:= \\ $$$$\frac{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}\mathrm{tan}\:^{\mathrm{2}} \theta}{\mathrm{tan}\:^{\mathrm{2}} \theta−\frac{\mathrm{1}}{\mathrm{3}}}\:=\:\frac{\mathrm{3}−\mathrm{tan}\:^{\mathrm{2}} \theta}{\mathrm{3tan}\:^{\mathrm{2}} \theta−\mathrm{1}}\::\:\left[\:\frac{\mathrm{tan}\:^{\mathrm{2}} \theta}{\mathrm{tan}\:^{\mathrm{2}} \theta}\:\right] \\ $$$$=\:\frac{\mathrm{3cot}\:^{\mathrm{2}} \theta−\mathrm{1}}{\mathrm{3}−\mathrm{cot}\:^{\mathrm{2}} \theta} \\ $$$$ \\ $$

Commented by bemath last updated on 01/Nov/20

(b) −3cot (θ+30°) cot (θ−30°)=cosec^2 θ  ⇒−3(((3cot^2 θ−1)/(3−cot^2 θ))) = 1+cot^2 θ  ⇒3−9cot^2 θ = (3−cot^2 θ)(1+cot^2 θ)  let cot^2 θ = s   ⇒3−9s = 3+2s−s^2   ⇒s^2 −11s = 0 → { ((s=0 (rejected))),((s=11)) :}  ⇒ cot θ = ± (√(11))  ⇒tan θ=±(1/( (√(11)))) ⇒θ = ± 16.77° +n.180°

$$\left(\mathrm{b}\right)\:−\mathrm{3cot}\:\left(\theta+\mathrm{30}°\right)\:\mathrm{cot}\:\left(\theta−\mathrm{30}°\right)=\mathrm{cosec}\:^{\mathrm{2}} \theta \\ $$$$\Rightarrow−\mathrm{3}\left(\frac{\mathrm{3cot}\:^{\mathrm{2}} \theta−\mathrm{1}}{\mathrm{3}−\mathrm{cot}\:^{\mathrm{2}} \theta}\right)\:=\:\mathrm{1}+\mathrm{cot}\:^{\mathrm{2}} \theta \\ $$$$\Rightarrow\mathrm{3}−\mathrm{9cot}\:^{\mathrm{2}} \theta\:=\:\left(\mathrm{3}−\mathrm{cot}\:^{\mathrm{2}} \theta\right)\left(\mathrm{1}+\mathrm{cot}\:^{\mathrm{2}} \theta\right) \\ $$$$\mathrm{let}\:\mathrm{cot}\:^{\mathrm{2}} \theta\:=\:\mathrm{s}\: \\ $$$$\Rightarrow\mathrm{3}−\mathrm{9s}\:=\:\mathrm{3}+\mathrm{2s}−\mathrm{s}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{s}^{\mathrm{2}} −\mathrm{11s}\:=\:\mathrm{0}\:\rightarrow\begin{cases}{\mathrm{s}=\mathrm{0}\:\left(\mathrm{rejected}\right)}\\{\mathrm{s}=\mathrm{11}}\end{cases} \\ $$$$\Rightarrow\:\mathrm{cot}\:\theta\:=\:\pm\:\sqrt{\mathrm{11}} \\ $$$$\Rightarrow\mathrm{tan}\:\theta=\pm\frac{\mathrm{1}}{\:\sqrt{\mathrm{11}}}\:\Rightarrow\theta\:=\:\pm\:\mathrm{16}.\mathrm{77}°\:+\mathrm{n}.\mathrm{180}° \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com