Maharashtra Board Solutions for Chapter: Trigonometric Functions, Exercise 9: Miscellaneous Exercise 3

Show that ${\tan }^{-1}\frac{1}{2}-{\tan }^{-1}\frac{1}{4}={\tan }^{-1}\frac{2}{9}$

Show that ${\cot }^{-1}\frac{1}{3}-{\tan }^{-1}\frac{1}{3}={\cot }^{-1}\frac{3}{4}$

Show that ${\cos }^{-1}\frac{\sqrt{3}}{2}+2{\sin }^{-1}\frac{\sqrt{3}}{2}=\frac{5\pi }{6}$

Show that $2{\cot }^{-1}\frac{3}{2}+{\sec }^{-1}\frac{13}{12}=\frac{\pi }{2}$

Prove the following :

${\cos }^{-1}\mathrm{x}={\tan }^{-1}\frac{\sqrt{1-x^2}}{x}$ if $\mathrm{x}<0$

Prove the following :

${\cos }^{-1}\mathrm{x}=\pi +{\tan }^{-1}\frac{\sqrt{1-x^2}}{x}$ if $\mathrm{x}<0$

If $\left|\mathrm{x}\right|<1$, then prove that $2{\tan }^{-1}x={\tan }^{-1}\frac{2x}{1-x^2}={\sin }^{-1}\frac{2x}{1+x^2}={\cos }^{-1}\frac{1-x^2}{1+x^2}$

If $x,y,z$ , are positive then prove that ${\tan }^{-1}\frac{x-y}{1+xy}+{\tan }^{-1}\frac{y-z}{1+yz}+{\tan }^{-1}\frac{z-x}{1+zx}=0$