Question

2cot inverse(7)+cos inverse (3/5)

Answer 1

$\textbf{Given:}$

$2\;cot^{-1}7+cos^{-1}\frac{3}{5}$

$\text{Take}\;2\;cot^{-1}7=A$

$\implies\;cot\frac{A}{2}=7$

$\implies\;tan\frac{A}{2}=\frac{1}{7}$

$\text{Using}$

$\bf\;tanA=\frac{2\;tan\frac{A}{2}}{1-tan^2\frac{A}{2}}$

$tanA=\frac{2(\frac{1}{7})}{1-\frac{1}{49}}$

$tanA=\frac{\frac{2}{7}}{\frac{48}{49}}$

$tanA=\frac{2}{7}{\times}\frac{49}{48}$

$tanA=\frac{7}{24}$

$\implies\boxed{A=tan^{-1}\frac{7}{24}}$

$\text{Take, }\;cos^{-1}\frac{3}{5}=B$

$\implies\;cosB=\frac{3}{5}$

$\implies\;secB=\frac{5}{3}$

$\text{Using}$

$\bf\;tan^2B=sec^2B-1$

$tan^2B=\frac{25}{9}-1$

$tan^2B=\frac{16}{9}$

$tanB=\frac{4}{3}$

$\implies\boxed{B=tan^{-1}\frac{4}{3}}$

$\text{Now,}$

$2\;cot^{-1}7+cos^{-1}\frac{3}{5}$

$=A+B$

$=tan^{-1}\frac{7}{24}+tan^{-1}\frac{4}{3}$

$\text{Using}$

$\boxed{tan^{-1}x+tan^{-1}y=tan^{-1}(\frac{x+y}{1-xy})}$

$=tan^{-1}(\frac{\frac{7}{24}+\frac{4}{3}}{1-\frac{7}{24}\frac{4}{3}})$

$=tan^{-1}(\frac{\frac{21+96}{72}}{1-\frac{28}{72}})$

$=tan^{-1}(\frac{\frac{117}{72}}{1-\frac{7}{18}})$

$=tan^{-1}(\frac{\frac{39}{24}}{\frac{11}{18}})$

$=tan^{-1}(\frac{\frac{39}{12}}{\frac{11}{9}})$

$=tan^{-1}(\frac{351}{132})$

$=tan^{-1}(\frac{117}{44})$