Question

Show that cos$\left(2ta{n}^{-1}\frac{1}{7}\right)=sin\left(4ta{n}^{-1}\frac{1}{3}\right).$

Answer 1

We have, cos$\left(2ta{n}^{-1}\frac{1}{7}\right)=sin\left(4ta{n}^{-1}\frac{1}{3}\right)\Rightarrow cos\left[co{s}^{-1}\left(\frac{1-{\left(\frac{1}{7}\right)}^2}{1+{\left(\frac{1}{7}\right)}^2}\right)\right]=sin\left[2.2ta{n}^{-1}\frac{1}{3}\right][\because 2ta{n}^{-1}x=co{s}^{-1}\left(\frac{1-x^2}{1+x^2}\right)]\Rightarrow cos\left[co{s}^{-1}\left(\frac{\frac{48}{49}}{\frac{50}{49}}\right)\right]=sin\left[2.\left(ta{n}^{-1}\frac{\frac{2}{3}}{1-{\left(\frac{1}{3}\right)}^2}\right)\right][\because 2ta{n}^{-1}x=ta{n}^{-1}\left(\frac{2x}{1-x^2}\right)]\Rightarrow cos\left[co{s}^{-1}\left(\frac{48\times 49}{50\times 49}\right)\right]=sin\left[2ta{n}^{-1}\left(\frac{18}{24}\right)\right]\Rightarrow cos\left[co{s}^{-1}\left(\frac{24}{25}\right)\right]=sin\left(2ta{n}^{-1}\frac{3}{4}\right)\Rightarrow cos\left[co{s}^{-1}\left(\frac{24}{25}\right)\right]=sin\left[si{n}^{-1}\frac{2\times \frac{3}{4}}{1+\frac{9}{16}}\right][\because 2ta{n}^{-1}x=si{n}^{-1}\frac{2x}{1+x^2}]\Rightarrow \frac{24}{25}=sin\left(si{n}^{-1}\frac{3/2}{25/16}\right)\Rightarrow \frac{24}{25}=\frac{48}{50}\Rightarrow \frac{24}{25}=\frac{24}{25}\therefore LHS=RHSHenceproved.$