NCERT Miscellaneous Exercise Solutions Chapter 2 Inverse Trigonometric Functions

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Monday, December 28, 2020

Agam Sir

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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}

Agam Sir

Agam Sir

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Ashish Kumar a.k.a Agam Sir has vast experience of 12+ years of teaching Mathematics and Physics. He has skills (academic and vocational) that go with working alongside people, especially young people, their parent(s) and the institutes that works with him. He always seek to be an agent of positive change and progression, a mentor/ trainer/ educator for learners/students, seeing that their needs are ideally met.