Class 12 Maths Inverse Trigonometric Functions worksheet 2 with answers

 Class 12 Maths Inverse Trigonometric Functions – Get here Important questions for Class 12 Maths Inverse Trigonometric Functions. Candidates who want to qualify the Class 12 with good score can practice these questions for the better understanding of concepts.These questions are framed as per the syllabus of the CBSE Board so go through these important questions for inverse trigonometric function with answers to get good score in class 12 maths CBSE board exam.

Inverse Trigonometric functions DPP – 2

$\begin{array}{l}1)Provethat\\ i){\sin }^{-1}\left(\frac{1}{x}\right)=\cos e{c}^{-1}x,\left(x\ge 1\text{or }x\le 1\right)\\ ii){\tan }^{-1}\left(\frac{1}{x}\right)={\cot }^{-1}x,\left(x>0\right)\\ 2)\text{ Find value of}\\ i){\sin }^{-1}\left(\sin \frac{\pi }{3}\right)\\ ii){\text{cos}}^{-1}\left(\cos \frac{2\pi }{3}\right)\\ iii){\text{ sin}}^{-1}\left(\sin \frac{2\pi }{3}\right)\\ iv){\text{ cos}}^{-1}\left(\cos \left(\frac{7\pi }{6}\right)\right)\\ 3)Show{\text{ that tan}}^{-1}\left(\tan \left(\frac{5\pi }{6}\right)\right)\ne \frac{5\pi }{6}\text{ and find its value}\text{.}\\ 4)\text{ Find value of}\\ i)\text{cos}\left({\tan }^{-1}\left(\frac{3}{4}\right)\right)\\ ii)\text{sin}\left\{\frac{\pi }{3}-{\sin }^{-1}\left(-\frac{1}{2}\right)\right\}\\ iii)\text{tan}\frac{1}{2}\left({\cos }^{-1}\frac{\sqrt{5}}{3}\right)\\ iv)\text{sin}\left[2{\cos }^{-1}\left(-\frac{3}{5}\right)\right]\\ 5)\text{ If }{\cot }^{-1}\left(-\frac{1}{5}\right)=\theta ,\text{ then find the value of }\theta \text{.}\\ 6){\text{ Prove that tan}}^{-1}\frac{3}{4}+{\tan }^{-1}\frac{3}{5}-{\tan }^{-1}\frac{8}{19}=\frac{\pi }{4}\\ 7){\text{Prove that tan}}^{-1}\frac{1}{3}+{\tan }^{-1}\frac{1}{5}+{\tan }^{-1}\frac{1}{7}+{\tan }^{-1}\frac{1}{8}=\frac{\pi }{4}\\ 8){\text{Prove that : 2tan}}^{-1}\frac{1}{2}+{\tan }^{-1}\frac{1}{7}={\tan }^{-1}\frac{31}{17}\\ 9){\text{ Prove that : cot}}^{-1}7+{\cot }^{-1}8+{\cot }^{-1}18={\cot }^{-1}3\\ 10){\text{ Prove that : tan}}^{-1}\frac{1}{4}+{\tan }^{-1}\frac{2}{9}=\frac{1}{2}{\cos }^{-1}\frac{3}{5}\end{array}$

ANSWERS $$\begin{array}{l}2)\\ \text{i)}\frac{\pi }{3}\text{ii) }\frac{2\pi }{3}\text{iii)}\frac{\pi }{3}\text{iv) }\frac{5\pi }{6}\\ 3)\left(-\frac{\pi }{6}\right)\\ 4)\\ \text{i) }\frac{4}{5}\text{ii)1}\text{iii)}\frac{3-\sqrt{5}}{2}\text{iv) }\frac{-24}{25}\\ 5)\frac{5}{\sqrt{26}}11)x=\frac{1}{6}12)x=\frac{1}{4}\end{array}$$

Other Resources :

•  Class 12 Maths Inverse Trigonometric Functions Important questions
• Class 12 Maths Inverse Trigonometry Functions NCERT Solutions