
Ch02. Inverse Trigonometric Functions
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In this online course, you will learn definition, range, domain, principal value branch, graphs of inverse trigonometric functions and elementary properties of inverse trigonometric functions. For further understanding of concepts and for examination preparation, practice questions based on the above topics are discussed in the form of assignments that have questions from NCERT Textbook exercise, NCERT Examples, Boardβs Question Bank, RD Sharma, NCERT Exemplar etc. instead of only one book. The PDF of assignments can be downloaded within the course.
Course Content
The following list of questions are just meant for reference before purchasing membership. The list might or might not include NCERT Questions as it depends on the chapter/course. Some chapters have NCERT questions combined in the Assignments and some chapters have separate NCERT questions and Assignments. For complete details, please check the index of the course in the "About Course".
Assignment β 1
Evaluate each of the following:
- {\displaystyle {{\sin }^{-1}}\left( -\frac{\sqrt{3}}{2} \right)}
- {\displaystyle {{\cot }^{-1}}\left( \frac{-1}{\sqrt{3}} \right)}
- {\displaystyle {{\tan }^{-1}}\left( -\frac{1}{\sqrt{3}} \right)}
- {\displaystyle {{\tan }^{-1}}(1)+{{\cos }^{-1}}\left( -\frac{1}{2} \right)+{{\sin }^{-1}}\left( -\frac{1}{2} \right)}
- {\displaystyle {{\cos }^{-1}}\left( \frac{1}{2} \right)+2{{\sin }^{-1}}\left( \frac{1}{2} \right)}
- {\displaystyle {{\tan }^{-1}}\sqrt{3}-{{\sec }^{-1}}(-2)}
- {\displaystyle {{\sin }^{-1}}\left( \sin \frac{4\pi }{5} \right)}
- {\displaystyle {{\sin }^{-1}}\left( \sin \frac{2\pi }{3} \right)}
- {\displaystyle {{\tan }^{-1}}\left( \tan \frac{3\pi }{4} \right)}
- {\displaystyle {{\cos }^{-1}}\left( \cos \frac{7\pi }{6} \right)}
- {\displaystyle \sin \left( \frac{\pi }{3}-{{\sin }^{-1}}\left( -\frac{1}{2} \right) \right)}
- {\displaystyle {{\tan }^{-1}}\sqrt{3}-{{\cot }^{-1}}(-\sqrt{3})}
- {\displaystyle {{\csc }^{-1}}(-2)}
- {\displaystyle {{\sin }^{-1}}\left( \sin \frac{3\pi }{5} \right)}
- {\displaystyle \tan \left[ \frac{1}{2}{{\cos }^{-1}}\left( \frac{2}{\sqrt{5}} \right) \right]}
- {\displaystyle {{\cos }^{-1}}\left( \cos \frac{13\pi }{6} \right)}
- {\displaystyle {{\tan }^{-1}}\left( \tan \frac{7\pi }{6} \right)}
- {\displaystyle {{\cos }^{-1}}[\cos (-680{}^\circ )]}
- {\displaystyle \tan \left\{ 2{{\tan }^{-1}}\frac{1}{5}-\frac{\pi }{4} \right\}}
- {\displaystyle {{\tan }^{-1}}\left( \tan \frac{5\pi }{6} \right)}
- {\displaystyle {{\sin }^{-1}}\left( \cos \left( \frac{43\pi }{5} \right) \right)}
- {\displaystyle {{\sin }^{-1}}\left( -\frac{\sqrt{3}}{2} \right)+{{\cos }^{-1}}\left( -\frac{1}{2} \right)+{{\tan }^{-1}}\left( -\frac{1}{\sqrt{3}} \right)}
- {\displaystyle {{\tan }^{2}}({{\sec }^{-1}}2)+{{\cot }^{2}}(\text{cose}{{\text{c}}^{-1}}3)}
- {\displaystyle \sin \left( 2{{\tan }^{-1}}\frac{1}{3} \right)+\cos ({{\tan }^{-1}}2\sqrt{2})}
- {\displaystyle {{\sec }^{2}}({{\tan }^{-1}}2)+\text{cose}{{\text{c}}^{2}}({{\cot }^{-1}}3)}
- {\displaystyle \sin ({{\tan }^{-1}}x+{{\cot }^{-1}}x)}
- {\displaystyle \sin \left( {{\cos }^{-1}}\frac{4}{5} \right)}
- {\displaystyle \sin \left( {{\cot }^{-1}}\frac{4}{3} \right)}
- {\displaystyle \sin ({{\cot }^{-1}}x)}
- {\displaystyle \cos ({{\tan }^{-1}}x)}
- {\displaystyle {{\sin }^{-1}}\frac{1}{2}-2{{\sin }^{-1}}\frac{1}{\sqrt{2}}}
- {\displaystyle \tan \left( {{\cos }^{-1}}\frac{4}{5}+{{\tan }^{-1}}\frac{2}{3} \right)}
- {\displaystyle \tan \left( 2{{\tan }^{-1}}\frac{1}{5} \right)}
- {\displaystyle \tan \left[ 2\cos \left( 2{{\sin }^{-1}}\frac{1}{2} \right) \right]}
- {\displaystyle \cot ({{\tan }^{-1}}a+{{\cot }^{-1}}a)}
- {\displaystyle \tan \frac{1}{2}\left[ {{\sin }^{-1}}\frac{2x}{1+{{x}^{2}}}+{{\cos }^{-1}}\frac{1-{{y}^{2}}}{1+{{y}^{2}}} \right],\, |x|<1,\,y>0\,\text{and}\,xy<1}
- {\displaystyle \sin ({{\tan }^{-1}}x),\,\,|x|<1}
- {\displaystyle {{\tan }^{-1}}\left( \frac{x}{y} \right)-{{\tan }^{-1}}\left( \frac{x-y}{x+y} \right)}
Assignment β 2
Simplify:
- {\displaystyle {{\tan }^{-1}}\left( \sqrt{\frac{1-\cos x}{1+\cos x}} \right),\,\,x<\pi }
- {\displaystyle {{\tan }^{-1}}\left( \frac{\cos x-\sin x}{\cos x+\sin x} \right),\,x<\pi }
- {\displaystyle {{\cot }^{-1}}\left( \frac{1}{\sqrt{{{x}^{2}}-1}} \right),\,\,|x|\,>1}
- {\displaystyle {{\tan }^{-1}}\frac{1}{\sqrt{{{x}^{2}}-1}},\,|x|\,>1}
- {\displaystyle {{\tan }^{-1}}\left( \frac{\sqrt{1+{{x}^{2}}}-1}{x} \right),\,\,x\ne 0}
- {\displaystyle {{\tan }^{-1}}\frac{x}{\sqrt{{{a}^{2}}-{{x}^{2}}}},\,\,|x|<a}
- {\displaystyle {{\tan }^{-1}}\left( \frac{3{{a}^{2}}x-{{x}^{3}}}{{{a}^{3}}-3a{{x}^{2}}} \right),\,a>0;\,\frac{-a}{\sqrt{3}}\le x\le \frac{a}{\sqrt{3}}}
- {\displaystyle {{\tan }^{-1}}\left( \frac{\cos x}{1+\sin x} \right)}
- {\displaystyle {{\tan }^{-1}}\left( \frac{\sin x}{1+\cos x} \right)}
- {\displaystyle {{\tan }^{-1}}\left( \frac{a\cos x-b\sin x}{b\cos x+a\sin x} \right)}
- {\displaystyle {{\tan }^{-1}}\left( \frac{x}{a+\sqrt{{{a}^{2}}-{{x}^{2}}}} \right)}
- {\displaystyle {{\sin }^{-1}}\left( \frac{5}{13}\cos x+\frac{12}{13}\sin x \right)}
- {\displaystyle {{\sin }^{-1}}(x\sqrt{1-x}-\sqrt{x}\sqrt{1-{{x}^{2}}})}
- {\displaystyle {{\sin }^{-1}}\left\{ \frac{\sqrt{1+x}+\sqrt{1-x}}{2} \right\}}
- {\displaystyle {{\tan }^{-1}}\sqrt{\frac{a-x}{a+x}}}
- {\displaystyle {{\tan }^{-1}}\left( \frac{\cos x}{1-\sin x} \right),\,\,-\frac{\pi }{2}<x<\frac{\pi }{2}}
- {\displaystyle {{\sin }^{-1}}\left\{ \frac{x+\sqrt{1-{{x}^{2}}}}{\sqrt{2}} \right\}}
- {\displaystyle {{\tan }^{-1}}(x+\sqrt{1+{{x}^{2}}})}
- {\displaystyle \sin \left\{ 2{{\tan }^{-1}}\sqrt{\frac{1-x}{1+x}} \right\}}
- {\displaystyle {{\sin }^{-1}}\left( \frac{\sin x+\cos x}{\sqrt{2}} \right)}
- {\displaystyle {{\tan }^{-1}}\left[ \frac{a\cos x-b\sin x}{b\cos x+a\sin x} \right],\,\text{if}\,\,\frac{a}{b}\tan x>-1}
- {\displaystyle {{\cot }^{-1}}\left( \frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}} \right)=\frac{x}{2},\,\,x\in \left( 0,\,\,\frac{\pi }{4} \right)}
- {\displaystyle {{\tan }^{-1}}\left( \frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}} \right)=\frac{\pi }{4}-\frac{1}{2}{{\cos }^{-1}}x,\,\,-\frac{1}{\sqrt{2}}\le x\le 1}
Assignment β 3
Prove:
- {\displaystyle 3{{\sin }^{-1}}x={{\sin }^{-1}}(3x-4{{x}^{3}}),\,\,x\in \left[ -\frac{1}{2},\frac{1}{2} \right]}
- {\displaystyle 3{{\cos }^{-1}}x={{\cos }^{-1}}(4{{x}^{3}}-3x),\,\,x\in \left[ \frac{1}{2},\,\,1 \right]}
- {\displaystyle {{\tan }^{-1}}\frac{2}{11}+{{\tan }^{-1}}\frac{7}{24}={{\tan }^{-1}}\frac{1}{2}}
- {\displaystyle 2{{\tan }^{-1}}\frac{1}{2}+{{\tan }^{-1}}\frac{1}{7}={{\tan }^{-1}}\frac{31}{17}}
- {\displaystyle {{\tan }^{-1}}\frac{3}{4}+{{\tan }^{-1}}\frac{3}{5}-{{\tan }^{-1}}\frac{8}{19}=\frac{\pi }{4}}
- {\displaystyle {{\cot }^{-1}}7+{{\cot }^{-1}}8+{{\cot }^{-1}}18={{\cot }^{-1}}3}
- {\displaystyle {{\tan }^{-1}}x+{{\tan }^{-1}}\left( \frac{2x}{1-{{x}^{2}}} \right)={{\tan }^{-1}}\left( \frac{3x-{{x}^{3}}}{1-3{{x}^{2}}} \right)}
- {\displaystyle {{\sin }^{-1}}(2x\sqrt{1-{{x}^{2}}})=2{{\sin }^{-1}}x=2{{\cos }^{-1}}x}
- {\displaystyle {{\tan }^{-1}}\left( \frac{\sqrt{1+{{x}^{2}}}+\sqrt{1-{{x}^{2}}}}{\sqrt{1+{{x}^{2}}}-\sqrt{1-{{x}^{2}}}} \right)=\frac{\text{ }\!\!\pi\!\!\text{ }}{2}-\frac{1}{2}{{\sin }^{-1}}{{x}^{2}}}
- {\displaystyle {{\tan }^{-1}}\sqrt{x}=\frac{1}{2}{{\cos }^{-1}}\left( \frac{1-x}{1+x} \right),\,\,x\in [0,\,1]}
- {\displaystyle \frac{9\pi }{8}-\frac{9}{4}{{\sin }^{-1}}\frac{1}{3}=\frac{9}{4}{{\sin }^{-1}}\frac{2\sqrt{2}}{3}}
- {\displaystyle \tan \left( \frac{\pi }{4}+\frac{1}{2}{{\cos }^{-1}}\frac{a}{b} \right)+\tan \left( \frac{\pi }{4}-\frac{1}{2}{{\cos }^{-1}}\frac{a}{b} \right)=\frac{2b}{a}}
- {\displaystyle \sin [{{\cot }^{-1}}\{\cos ({{\tan }^{-1}}x)\}]=\frac{\sqrt{{{x}^{2}}+1}}{\sqrt{{{x}^{2}}+2}}}
- {\displaystyle {{\tan }^{-1}}\left( \frac{x}{\sqrt{{{a}^{2}}-{{x}^{2}}}} \right)={{\sin }^{-1}}\frac{x}{a}={{\cot }^{-1}}\left( \frac{\sqrt{{{a}^{2}}-{{x}^{2}}}}{a} \right)}
- {\displaystyle {{\tan }^{-1}}\left( \frac{m}{n} \right)-{{\tan }^{-1}}\left( \frac{m-n}{m+n} \right)=\frac{\pi }{4}}
- {\displaystyle {{\tan }^{-1}}\frac{1}{5}+{{\tan }^{-1}}\frac{1}{7}+{{\tan }^{-1}}\frac{1}{3}+{{\tan }^{-1}}\frac{1}{8}=\frac{\pi }{4}}
- {\displaystyle 4{{\tan }^{-1}}\frac{1}{5}-{{\tan }^{-1}}\frac{1}{70}+{{\tan }^{-1}}\frac{1}{99}=\frac{\pi }{4}}
- {\displaystyle {{\tan }^{-1}}\left( \frac{\cos x}{1-\sin x} \right)-{{\cot }^{-1}}\left( \sqrt{\frac{1+\cos x}{1-\cos x}} \right)=\frac{\pi }{4}}
- {\displaystyle {{\sin }^{-1}}\frac{3}{5}-{{\sin }^{-1}}\frac{8}{17}={{\cos }^{-1}}\frac{84}{85}}
- {\displaystyle {{\sin }^{-1}}\frac{8}{17}+{{\sin }^{-1}}\frac{3}{5}={{\tan }^{-1}}\frac{77}{36}}
- {\displaystyle {{\tan }^{-1}}\frac{63}{16}={{\sin }^{-1}}\frac{5}{13}+{{\cos }^{-1}}\frac{3}{5}}
- {\displaystyle {{\cos }^{-1}}\frac{4}{5}+{{\cos }^{-1}}\frac{12}{13}={{\cos }^{-1}}\frac{33}{65}}
- {\displaystyle {{\cos }^{-1}}\frac{12}{13}+{{\sin }^{-1}}\frac{3}{5}={{\sin }^{-1}}\frac{56}{65}}
- {\displaystyle {{\sin }^{-1}}\frac{12}{13}+{{\cos }^{-1}}\frac{4}{5}+{{\tan }^{-1}}\frac{63}{16}=\pi }
- {\displaystyle 2{{\tan }^{-1}}\left\{ \tan \frac{\alpha }{2}\tan \left( \frac{\pi }{4}-\frac{\beta }{2} \right) \right\}={{\tan }^{-1}}\frac{\sin \alpha \cos \beta }{\cos \alpha +\sin \beta }}
- {\displaystyle 2{{\tan }^{-1}}\left( \sqrt{\frac{a-b}{a+b}}\tan \frac{\theta }{2} \right)={{\cos }^{-1}}\left( \frac{a\cos \theta +b}{a+b\cos \theta } \right)}
- If {\displaystyle \sin \left( {{\sin }^{-1}}\frac{1}{5}+{{\cos }^{-1}}x \right)=1} , then find the value of x.
- If {\displaystyle y={{\cot }^{-1}}(\sqrt{\cos x})-{{\tan }^{-1}}(\sqrt{\cos x})} , prove that {\displaystyle \sin y={{\tan }^{2}}\frac{x}{2}} .
Assignment β 4
Solve:
- {\displaystyle {{\tan }^{-1}}\frac{x-1}{x-2}+{{\tan }^{-1}}\frac{x+1}{x+2}=\frac{\pi }{4}}
- {\displaystyle {{\tan }^{-1}}2x+{{\tan }^{-1}}3x=\frac{\pi }{4}}
- {\displaystyle {{\tan }^{-1}}(x+1)+{{\tan }^{-1}}(x-1)={{\tan }^{-1}}\frac{8}{31}}
- {\displaystyle 2{{\tan }^{-1}}(\cos x)={{\tan }^{-1}}(2\,\text{cosec}\,x)}
- {\displaystyle {{\tan }^{-1}}\frac{1-x}{1+x}=\frac{1}{2}{{\tan }^{-1}}x,\,(x>0)}
- {\displaystyle {{\sin }^{-1}}(1-x)-2{{\sin }^{-1}}x=\frac{\pi }{2}}
- {\displaystyle {{\sin }^{-1}}x+{{\sin }^{-1}}(1-x)={{\cos }^{-1}}x}
- {\displaystyle {{\sin }^{-1}}6x+{{\sin }^{-1}}6\sqrt{3}x=-\frac{\pi }{2}}
- Solve: {\displaystyle {{\tan }^{-1}}(x-1)+{{\tan }^{-1}}x+{{\tan }^{-1}}(x+1)={{\tan }^{-1}}3x}
- Solve: {\displaystyle 3{{\sin }^{-1}}\frac{2x}{1+{{x}^{2}}}-4{{\cos }^{-1}}\frac{1-{{x}^{2}}}{1+{{x}^{2}}}+2{{\tan }^{-1}}\frac{2x}{1-{{x}^{2}}}=\frac{\pi }{3}}
- If {\displaystyle {{\sin }^{-1}}\frac{2a}{1+{{a}^{2}}}-{{\cos }^{-1}}\frac{1-{{b}^{2}}}{1+{{b}^{2}}}={{\tan }^{-1}}\frac{2x}{1-{{x}^{2}}}} , then prove that {\displaystyle x=\frac{a-b}{1+ab}} .
- Evaluate: {\displaystyle {{\tan }^{-1}}\left( \frac{a+bx}{b-ax} \right),\,\,x<\frac{b}{a}}
- Prove: {\displaystyle {{\tan }^{-1}}\left( \frac{a-b}{1+ab} \right)+{{\tan }^{-1}}\left( \frac{b-c}{1+bc} \right)+{{\tan }^{-1}}\left( \frac{c-a}{1+ca} \right)=0}
- If {\displaystyle {{\tan }^{-1}}x+{{\tan }^{-1}}y=\frac{4\pi }{5}} , then find the value of {\displaystyle {{\cot }^{-1}}x+{{\cot }^{-1}}y} ?
- If {\displaystyle {{\tan }^{-1}}\left( \frac{1}{1+1.2} \right)+{{\tan }^{-1}}\left( \frac{1}{1+2.3} \right)+...+{{\tan }^{-1}}\left( \frac{1}{1+n.(n+1)} \right)={{\tan }^{-1}}\phi } , then find the value of {\displaystyle \phi } .
- If {\displaystyle {{({{\tan }^{-1}}x)}^{2}}+{{({{\cot }^{-1}}x)}^{2}}=\frac{5{{\pi }^{2}}}{8}} , then find {\displaystyle x} .
Syllabus medium | English |
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Explanation Language | Hinglish (Hindi + English) |
Class | 12 |
Course Mode | Online learning |
Learning mode | Self-learning from videos |
Subject | Mathematics |