Question Number 69953 by Shamim last updated on 29/Sep/19
$$\mathrm{if}\:\mathrm{tan}\:\theta+\mathrm{sec}\:\theta=\:\sqrt{\mathrm{3}}\:\mathrm{than}\:\mathrm{find}\:\mathrm{out}\:\mathrm{the} \\ $$$$\mathrm{value}\:\mathrm{of}\underset{} {\:}\theta\:\mathrm{where}\:\mathrm{0}^{\mathrm{o}} \leqslant\theta\leqslant\mathrm{2}\pi. \\ $$
Commented by Shamim last updated on 29/Sep/19
$$\mathrm{ans}\:\mathrm{ki}\:\mathrm{only}\:\mathrm{x}=\frac{\pi}{\mathrm{6}}??? \\ $$
Commented by mathmax by abdo last updated on 29/Sep/19
![tanθ +(1/(cosθ)) =(√3) ⇒cosθ tanθ +1=(√3)cosθ ⇒ sinθ +1=(√3)cosθ ⇒(√3)cosθ −sinθ =1 (with θ≠(π/2)+kπ) 2(((√3)/2)cosθ −(1/2)sinθ)=1 ⇒cosθ cos((π/6))−sinθ sin((π/6))=(1/2) ⇒ cos(θ+(π/6))=(1/2) ⇒cos(θ+(π/6))=cos((π/3)) ⇒θ+(π/6) =(π/3) +2kπ or θ+(π/6) =−(π/3) +2kπ ⇒θ =(π/6) +2kπ or θ =−(π/2) +2kπ (k∈Z) solutions on [0,2π] 0≤(π/6)+2kπ≤2π ⇒0≤(1/6) +2k ≤2 ⇒−(1/6)≤2k≤((11)/6) ⇒ −(1/(12))≤k≤((11)/(12)) ⇒k=0 ⇒x=(π/6) c_2 →0≤−(π/2)+2kπ≤2π ⇒0 ≤−(1/2) +2k≤2 ⇒(1/2) ≤2k≤(5/2) ⇒ (1/4)≤k≤(5/4) ⇒k=1 ⇒x =((3π)/2) but this is not so<ution!](https://www.tinkutara.com/question/Q69959.png)
$${tan}\theta\:+\frac{\mathrm{1}}{{cos}\theta}\:=\sqrt{\mathrm{3}}\:\Rightarrow{cos}\theta\:{tan}\theta\:+\mathrm{1}=\sqrt{\mathrm{3}}{cos}\theta\:\Rightarrow \\ $$$${sin}\theta\:+\mathrm{1}=\sqrt{\mathrm{3}}{cos}\theta\:\:\Rightarrow\sqrt{\mathrm{3}}{cos}\theta\:−{sin}\theta\:=\mathrm{1}\:\left({with}\:\theta\neq\frac{\pi}{\mathrm{2}}+{k}\pi\right) \\ $$$$\mathrm{2}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{cos}\theta\:−\frac{\mathrm{1}}{\mathrm{2}}{sin}\theta\right)=\mathrm{1}\:\Rightarrow{cos}\theta\:{cos}\left(\frac{\pi}{\mathrm{6}}\right)−{sin}\theta\:{sin}\left(\frac{\pi}{\mathrm{6}}\right)=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow \\ $$$${cos}\left(\theta+\frac{\pi}{\mathrm{6}}\right)=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow{cos}\left(\theta+\frac{\pi}{\mathrm{6}}\right)={cos}\left(\frac{\pi}{\mathrm{3}}\right)\:\Rightarrow\theta+\frac{\pi}{\mathrm{6}}\:=\frac{\pi}{\mathrm{3}}\:+\mathrm{2}{k}\pi\:{or} \\ $$$$\theta+\frac{\pi}{\mathrm{6}}\:=−\frac{\pi}{\mathrm{3}}\:+\mathrm{2}{k}\pi\:\Rightarrow\theta\:=\frac{\pi}{\mathrm{6}}\:+\mathrm{2}{k}\pi\:{or}\:\theta\:=−\frac{\pi}{\mathrm{2}}\:+\mathrm{2}{k}\pi\:\:\:\left({k}\in{Z}\right) \\ $$$${solutions}\:{on}\:\left[\mathrm{0},\mathrm{2}\pi\right] \\ $$$$\mathrm{0}\leqslant\frac{\pi}{\mathrm{6}}+\mathrm{2}{k}\pi\leqslant\mathrm{2}\pi\:\Rightarrow\mathrm{0}\leqslant\frac{\mathrm{1}}{\mathrm{6}}\:+\mathrm{2}{k}\:\leqslant\mathrm{2}\:\Rightarrow−\frac{\mathrm{1}}{\mathrm{6}}\leqslant\mathrm{2}{k}\leqslant\frac{\mathrm{11}}{\mathrm{6}}\:\Rightarrow \\ $$$$−\frac{\mathrm{1}}{\mathrm{12}}\leqslant{k}\leqslant\frac{\mathrm{11}}{\mathrm{12}}\:\Rightarrow{k}=\mathrm{0}\:\Rightarrow{x}=\frac{\pi}{\mathrm{6}} \\ $$$${c}_{\mathrm{2}} \rightarrow\mathrm{0}\leqslant−\frac{\pi}{\mathrm{2}}+\mathrm{2}{k}\pi\leqslant\mathrm{2}\pi\:\Rightarrow\mathrm{0}\:\leqslant−\frac{\mathrm{1}}{\mathrm{2}}\:+\mathrm{2}{k}\leqslant\mathrm{2}\:\Rightarrow\frac{\mathrm{1}}{\mathrm{2}}\:\leqslant\mathrm{2}{k}\leqslant\frac{\mathrm{5}}{\mathrm{2}}\:\Rightarrow \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\leqslant{k}\leqslant\frac{\mathrm{5}}{\mathrm{4}}\:\Rightarrow{k}=\mathrm{1}\:\Rightarrow{x}\:=\frac{\mathrm{3}\pi}{\mathrm{2}}\:\:{but}\:{this}\:{is}\:{not}\:{so}<{ution}! \\ $$$$ \\ $$$$ \\ $$
Commented by Prithwish sen last updated on 29/Sep/19
$$\mathrm{tan}\theta+\mathrm{sec}\theta\:=\:\sqrt{\mathrm{3}}….\left(\mathrm{i}\right) \\ $$$$\Rightarrow\mathrm{sec}\theta−\mathrm{tan}\theta\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}…\left(\mathrm{ii}\right) \\ $$$$\mathrm{Adding}\:\left(\mathrm{i}\right)\:\mathrm{and}\:\left(\mathrm{ii}\right) \\ $$$$\mathrm{2sec}\theta=\frac{\mathrm{4}}{\:\sqrt{\mathrm{3}}} \\ $$$$\mathrm{cos}\theta=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}=\mathrm{cos}\frac{\pi}{\mathrm{6}} \\ $$$$\Rightarrow\:\boldsymbol{\theta}\:=\:\boldsymbol{\mathrm{n}\pi}\:\pm\:\frac{\boldsymbol{\pi}}{\mathrm{6}} \\ $$
Commented by mathmax by abdo last updated on 29/Sep/19
$${yes}. \\ $$
Commented by Shamim last updated on 29/Sep/19
$$\mathrm{okk}. \\ $$$$\mathrm{all}\:\mathrm{kind}\:\mathrm{of}\:\mathrm{like}\:\mathrm{this}\:\mathrm{math}\:\mathrm{maintain} \\ $$$$\mathrm{same}\:\mathrm{rule}??? \\ $$