Question and Answers Forum

All Questions      Topic List

Trigonometry Questions

Previous in All Question      Next in All Question      

Previous in Trigonometry      Next in Trigonometry      

Question Number 51364 by rahul 19 last updated on 26/Dec/18

The value of x for which sin(cot^(−1) (1+x))=cos(tan^(−1) x) is ?

$${The}\:{value}\:{of}\:{x}\:{for}\:{which} \\ $$$$\mathrm{sin}\left(\mathrm{cot}^{−\mathrm{1}} \left(\mathrm{1}+{x}\right)\right)=\mathrm{cos}\left(\mathrm{tan}^{−\mathrm{1}} {x}\right)\:{is}\:? \\ $$

Commented by rahul 19 last updated on 26/Dec/18

sin(cot^(−1) (1+x))=sin((π/2)−tan^(−1) x) ⇒tan^(−1) ((1/(1+x)))=((π/2)−tan^(−1) x) ⇒tan^(−1) ((1/(1+x)))+tan^(−1) x=(π/2) ⇒tan^(−1) ((((1/(1+x))+x)/(1−(x/(1+x)))))= (π/2) ⇒tan^(−1) (((x^2 +x+1)/1))=(π/2) ..... ?????

$${sin}\left(\mathrm{cot}^{−\mathrm{1}} \left(\mathrm{1}+{x}\right)\right)=\mathrm{sin}\left(\frac{\pi}{\mathrm{2}}−\mathrm{tan}^{−\mathrm{1}} {x}\right) \\ $$$$\Rightarrow\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{1}+{x}}\right)=\left(\frac{\pi}{\mathrm{2}}−\mathrm{tan}^{−\mathrm{1}} {x}\right) \\ $$$$\Rightarrow{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{1}+{x}}\right)+\mathrm{tan}^{−\mathrm{1}} {x}=\frac{\pi}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{tan}^{−\mathrm{1}} \left(\frac{\frac{\mathrm{1}}{\mathrm{1}+{x}}+{x}}{\mathrm{1}−\frac{{x}}{\mathrm{1}+{x}}}\right)=\:\frac{\pi}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{tan}^{−\mathrm{1}} \left(\frac{{x}^{\mathrm{2}} +{x}+\mathrm{1}}{\mathrm{1}}\right)=\frac{\pi}{\mathrm{2}} \\ $$$$.....\:????? \\ $$

Commented by prakash jain last updated on 26/Dec/18

cot^(−1) (1+x)=θ cot θ=1+x sin θ=(1/(√(1+(1+x)^2 ))) cos(tan^(−1) x)=(1/(√(1+x^2 ))) 1+(1+x)^2 =1+x^2 2x=−1⇒x=−(1/2)

$$\mathrm{cot}^{−\mathrm{1}} \left(\mathrm{1}+{x}\right)=\theta \\ $$$$\mathrm{cot}\:\theta=\mathrm{1}+{x} \\ $$$$\mathrm{sin}\:\theta=\frac{\mathrm{1}}{\sqrt{\mathrm{1}+\left(\mathrm{1}+{x}\right)^{\mathrm{2}} }} \\ $$$$\mathrm{cos}\left(\mathrm{tan}^{−\mathrm{1}} {x}\right)=\frac{\mathrm{1}}{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }} \\ $$$$\mathrm{1}+\left(\mathrm{1}+{x}\right)^{\mathrm{2}} =\mathrm{1}+{x}^{\mathrm{2}} \\ $$$$\mathrm{2}{x}=−\mathrm{1}\Rightarrow{x}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com