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 64773 by ankan0 last updated on 21/Jul/19

$${\sin }^{-1}(1/\sqrt{5)}+{\cot }^{-1}3$$

Answered by Kunal12588 last updated on 21/Jul/19

$${\cot }^{-1}\left(a\right)={\sin }^{-1}\left(\frac{1}{\sqrt{1+a^2}}\right)$$$${\cot }^{-1}3={\sin }^{-1}\left(\frac{1}{\sqrt{1+9}}\right)={\sin }^{-1}\frac{1}{\sqrt{10}}$$$${\sin }^{-1}a+{\sin }^{-1}b={\sin }^{-1}(a\sqrt{1-b^2}+b\sqrt{1-a^2})$$$${\sin }^{-1}\left(1/\sqrt{5}\right)+{\cot }^{-1}3$$$$={\sin }^{-1}\frac{1}{\sqrt{5}}+{\sin }^{-1}\frac{1}{\sqrt{10}}$$$$={\sin }^{-1}(\frac{1}{\sqrt{5}}\sqrt{1-\frac{1}{10}}+\frac{1}{\sqrt{10}}\sqrt{1-\frac{1}{5}})$$$$={\sin }^{-1}(\frac{3}{\sqrt{5}\sqrt{10}}+\frac{2}{\sqrt{10}\sqrt{5}})$$$$={\sin }^{-1}\left(\frac{5}{5\sqrt{2}}\right)$$$$={\sin }^{-1}\left(\frac{1}{\sqrt{2}}\right)$$$$=\frac{\pi }{4}=45°$$

Commented by John M. Kaloki last updated on 13/Aug/19

$$Ican$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com