NCERT Miscellaneous Exercise Solutions Chapter 2 Inverse Trigonometric Functions

by | Dec 28, 2020

Topics/Questions discussed in this video:

NCERT Miscellaneous Exercise Class 12 Maths Chapter 2 Inverse Trigonometric Functions

Question-1. \cos^{-1}\cos⁡\frac{13\pi}{6}  

Question-2. \tan^{-1}⁡\tan\frac⁡{7\pi}{6}

Question-3. 2 \sin^{-1}\frac{3}{5}= \tan^{-1}\frac{24}{7}

Question-4. \sin^{-1}⁡\frac{8}{17}+\sin^{-1}\frac{⁡3}{5}=\tan^{-1}\frac{⁡77}{36}

Question-5. \cos^{-1}\frac{⁡4}{5}+\cos^{-1}\frac{⁡12}{13}=\cos^{-1}\frac{⁡33}{65} 

Question-6. \cos^{-1}\frac{⁡12}{13}+\sin^{-1}\frac{⁡3}{5}=\sin^{-1}\frac{⁡56}{65} 

Question-7. \tan^{-1}\frac{⁡63}{16}=\sin^{-1}⁡\frac{5}{13}+\cos^{-1}\frac{⁡3}{5} 

Question-8. \tan^{-1}\frac{⁡1}{5}+\tan^{-1}\frac{⁡1}{7}+\tan^{-1}⁡\frac{1}{3}+\tan^{-1}⁡\frac{1}{8}=\frac{\pi}{4} 

Question-9. \tan^{-1}⁡\sqrt{x}=\frac{1}{2} \cos^{-1}⁡\frac{1-x}{1+x},x \in 0,1 

Question-10. cot^{-1}⁡\left ( \frac{\sqrt{1+\sin ⁡x }+\sqrt{1-\sin⁡x }}{\sqrt{1+\sin⁡ x }-\sqrt{1-\sin⁡x}} \right )=\frac{x}{2}

Question-11. \tan^{-1}⁡\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}=\frac{\pi}{4}-\frac{1}{2} \cos^{-1}⁡x

Question-12. \frac{9\pi}{8}-\frac{9}{4} \sin^{-1}\frac{⁡1}{3}=\frac{9}{4} \sin^{-1}⁡\frac{2\sqrt{2}}{3} 

Question-13. 2 \tan^{-1}⁡{ \cos⁡x}=\tan^{-1}⁡{ 2\cosec x} 

Question-14. \tan^{-1}⁡\frac{1-x}{1+x}=\frac{1}{2} \tan^{-1}⁡x,

Question-15. \sin⁡(\tan^{-1}⁡x),

Question-16. \sin^{-1}(1-x)-2 \sin^{-1}⁡x=\frac{\pi}{2} 

Question-17. \tan^{-1}\frac⁡{x}{y} -\tan^{-1}⁡\frac{x-y}{x+y}