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Question Number 94344 by i jagooll last updated on 18/May/20
if tan^(−1) (x)=(1/2)cos^(−1) ((5/(13)))  find x
$$\mathrm{if}\:\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}^{−\mathrm{1}} \left(\frac{\mathrm{5}}{\mathrm{13}}\right) \\ $$$$\mathrm{find}\:\mathrm{x}\: \\ $$
Answered by john santu last updated on 18/May/20
let θ = cos^(−1) ((5/(13))) ⇒ cos θ = (5/(13))  ⇒tan^(−1) (x)=(θ/2) ⇒x = tan ((θ/2))  remember tan θ= ((2tan ((θ/2)))/(1−tan^2  ((θ/2))))  ((12)/5) = ((2x)/(1−x^2 )) ⇒6−6x^2 = 5x  6x^2 +5x−6 = 0  (3x−2)(2x+3) = 0 ⇒ { ((x=(2/3))),((x=−(3/2))) :}
$$\mathrm{let}\:\theta\:=\:\mathrm{cos}^{−\mathrm{1}} \left(\frac{\mathrm{5}}{\mathrm{13}}\right)\:\Rightarrow\:\mathrm{cos}\:\theta\:=\:\frac{\mathrm{5}}{\mathrm{13}} \\ $$$$\Rightarrow\mathrm{tan}^{−\mathrm{1}} \left({x}\right)=\frac{\theta}{\mathrm{2}}\:\Rightarrow{x}\:=\:\mathrm{tan}\:\left(\frac{\theta}{\mathrm{2}}\right) \\ $$$$\mathrm{remember}\:\mathrm{tan}\:\theta=\:\frac{\mathrm{2tan}\:\left(\frac{\theta}{\mathrm{2}}\right)}{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \:\left(\frac{\theta}{\mathrm{2}}\right)} \\ $$$$\frac{\mathrm{12}}{\mathrm{5}}\:=\:\frac{\mathrm{2}{x}}{\mathrm{1}−{x}^{\mathrm{2}} }\:\Rightarrow\mathrm{6}−\mathrm{6}{x}^{\mathrm{2}} =\:\mathrm{5}{x} \\ $$$$\mathrm{6}{x}^{\mathrm{2}} +\mathrm{5}{x}−\mathrm{6}\:=\:\mathrm{0} \\ $$$$\left(\mathrm{3}{x}−\mathrm{2}\right)\left(\mathrm{2}{x}+\mathrm{3}\right)\:=\:\mathrm{0}\:\Rightarrow\begin{cases}{{x}=\frac{\mathrm{2}}{\mathrm{3}}}\\{{x}=−\frac{\mathrm{3}}{\mathrm{2}}}\end{cases} \\ $$$$ \\ $$
Commented by i jagooll last updated on 18/May/20
thank you
$$\mathrm{thank}\:\mathrm{you} \\ $$

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