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 140494 by benjo_mathlover last updated on 08/May/21

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

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

Answered by john_santu last updated on 08/May/21

⇒ tan^(−1) (((1−x)/(1+x)))=(π/2)−cot^(−1) (((1+x)/(1−x))) ⇒tan (tan^(−1) (((1−x)/(1+x))))= tan ((π/2)−cot^(−1) (((1+x)/(1−x)))) ⇒ ((1−x)/(1+x)) = cot (cot^(−1) (((1+x)/(1−x)))) ⇒((1−x)/(1+x)) = ((1+x)/(1−x)) ⇒(1−x)^2 =(1+x)^2 ⇒1−2x+x^2 = 1+2x+x^2 ⇒ x = 0

$$\Rightarrow\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\right)=\frac{\pi}{\mathrm{2}}−\mathrm{cot}^{−\mathrm{1}} \left(\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}\right) \\ $$$$\Rightarrow\mathrm{tan}\:\left(\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\right)\right)=\:\mathrm{tan}\:\left(\frac{\pi}{\mathrm{2}}−\mathrm{cot}^{−\mathrm{1}} \left(\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}\right)\right) \\ $$$$\Rightarrow\:\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\:=\:\mathrm{cot}\:\left(\mathrm{cot}^{−\mathrm{1}} \left(\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}\right)\right) \\ $$$$\Rightarrow\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\:=\:\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}\:\Rightarrow\left(\mathrm{1}−{x}\right)^{\mathrm{2}} =\left(\mathrm{1}+{x}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{1}−\mathrm{2}{x}+{x}^{\mathrm{2}} \:=\:\mathrm{1}+\mathrm{2}{x}+{x}^{\mathrm{2}} \\ $$$$\Rightarrow\:{x}\:=\:\mathrm{0} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com