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CBSE Assignments class 09 Mathematics

  Mathematics Assignments & Worksheets  For  Class IX Chapter-wise mathematics assignment for class 09. Important and useful extra questions strictly according to the CBSE syllabus and pattern with answer key CBSE Mathematics is a very good platform for the students and is contain the assignments for the students from 9 th  to 12 th  standard.  Here students can find very useful content which is very helpful to handle final examinations effectively.  For better understanding of the topic students should revise NCERT book with all examples and then start solving the chapter-wise assignments.  These assignments cover all the topics and are strictly according to the CBSE syllabus.  With the help of these assignments students can easily achieve the examination level and  can reach at the maximum height. Class 09 Mathematics    Assignment Case Study Based Questions Class IX 

Math Assignment Class XII Ch-2 : I T F

Maths Assignment | Class 12 | Chapter - 2 
Inverse Trigonometric Functions

Important questions on Inverse Trigonometric Functions, Maths Assignment on trigonometric functions, extra questions on inverse trigonometric functions class XII chapter 2

Important steps that students need to do
  • First of all students should learn and  revise all basic concepts and formulas of Inverse Trigonometric Functions.
  • Revise Chapter 2 Inverse Trigonometric Functions from  NCERT book of Mathematics.
  • Revise all examples of chapter 2 of  NCERT book.
  • Now start the Assignment.
Assignment on Inverse Trigonometric Functions

 Question 1: Find the Principal Value of the following functions

\[(i)\:\: \: \: \: cos^{-1}\left ( cos\frac{7\pi }{6} \right )\: \: .......\: \: Ans\:\: \: \frac{5\pi }{6}\]

\[(ii)\:\: \: \: \: cos^{-1}\left ( cos\frac{2\pi }{3} \right )+sin^{-1}\left ( sin\frac{2\pi }{3} \right )\: \: .......\: \: Ans\:\: \: (\pi )\]

\[(iii)\:\: \: \: \: tan^{-1}\left ( tan\frac{2\pi }{3} \right )+tan^{-1}\left ( tan\frac{9\pi }{8} \right )\: \: .......\: \: Ans\:\: \:\left (\frac{-5\pi }{24} \right )\]
\[(iv)\:\: \: \: \: sec^{-1}\left ( 2sin\frac{3\pi }{4} \right )\: \: .......\: \: Ans\:\: \:\left (\frac{\pi }{4} \right )\]

Question 2: \[Find \; \; the \: value \: of\: \: \: 4[cot^{-1}(3)+cosec^{-1}(\sqrt{5})]\: \: .......\: \: Ans \; = \pi\]

Question 3: Find the value of the following

\[2sin^{-1}\frac{1}{2}+3tan^{-1}(-1)+2cos^{-1}\left ( \frac{-1}{2} \right )+4sec^{-1}(\sqrt{2}) \\\: \: .......\: \: Ans\:\: \:\left (\frac{23\pi }{12} \right )\]
Question 4: Find the value of the following
\[4\left [ 2sin^{-1}\left ( \frac{-1}{2} \right )+5tan^{-1}(1)-3cos^{-1}\left ( \frac{1}{2} \right ) \right ]+\frac{1}{2}cos^{-1}\left ( \frac{-\sqrt{3}}{2} \right )\\ \: \: .......\: \: Ans\:\: \:\left (\frac{\pi }{12} \right )\]
Question 5: Find the value of the following
\[4cos^{-1}\left ( \frac{1}{2} \right )+2sin^{-1}\left ( \frac{1}{2} \right )+3tan^{-1}(-1)+2cos^{-1}\left ( \frac{\sqrt{3}}{2} \right )\\\: .........\: Ans\left ( \frac{5\pi }{4} \right )\]

 Question 6: Find the value of the following

\[sin^{-1}\left ( sin\frac{3\pi }{5} \right )+sin^{-1}\left ( sin\frac{4\pi }{5} \right )\; \; .......\; \; Ans\left ( \frac{3\pi }{5} \right )\]

 Question 7: Write the following functions in the simplest form

\[(i)\; \; \; tan^{-1}\left ( \sqrt{\frac{a-x}{a+x}} \right )\; \; ........\; \; Ans:\; \left ( \frac{1}{2}cos^{-1}\frac{x}{a} \right )\]
\[(ii)\; \; \; cos^{-1}\left ( \frac{x}{\sqrt{x^{2}+a^{2}}} \right )\; \; ........\; \; Ans:\; cot^{-1}\left (\frac{x}{a} \right )\]
\[(iii)\; \; \; sin^{-1}\left ( \frac{x}{\sqrt{x^{2}+a^{2}}} \right )\; \; ........\; \; Ans:\; tan^{-1}\frac{x}{a}\]
\[(iv)\; \; \; sin\left ( 2sin^{-1}\frac{3}{5} \right ) \; ........\; \; Ans:\; \frac{24}{25}\]
\[(v)\; \; \; tan^{-1}\left [ 2sin\left ( 2cos^{-1}\left ( \frac{\sqrt{3}}{2} \right ) \right ) \right ] \; ........\; \; Ans:\; \frac{\pi }{3}\]

Question 8: Simplify the followings

\[(i)\; \; \; sin^{-1}\left ( \frac{sinx+cosx}{\sqrt{2}} \right ) \; ........\; \; Ans:\; x+\frac{\pi }{4}\]\[(ii)\; \; \; cos^{-1}\left ( \frac{sinx+cosx}{\sqrt{2}} \right ) \; ........\; \; Ans:\; x-\frac{\pi }{4}\]

Question 9: Find the value of x if 

\[2tan^{-1}(sinx)=tan^{-1}(2secx)\; \; .......\; \; Ans:\; \; x=\frac{\pi }{4}\]

 Question 10: Prove that

\[tan^{-1}\left ( \frac{1}{2} \right )+tan^{-1}\left ( \frac{2}{11} \right )=tan^{-1}\left ( \frac{3}{4} \right )=\frac{1}{2}sin^{-1}\frac{24}{25}\]
************************************

 Question 11: Prove that 

\[tan^{-1}\left ( \frac{1}{4} \right )+tan^{-1}\left ( \frac{2}{9} \right )=\frac{1}{2}cos^{-1}\left ( \frac{3}{5} \right )=sin^{-1}\frac{1}{\sqrt{5}}\]
Solution:
\[tan^{-1}\left ( \frac{1}{4} \right )+tan^{-1}\left ( \frac{2}{9} \right )=tan^{-1}\left ( \frac{\frac{1}{4}+\frac{2}{9}}{1-\frac{1}{4}\times \frac{2}{9}} \right )\]
\[=tan^{-1}\left ( \frac{17/36}{34/36} \right )=tan^{-1}\left ( \frac{1}{2} \right ) ............ (1)\]
\[Now\: let\:\; \: \frac{1}{2}cos^{-1}\left ( \frac{3}{5} \right )=\theta \Rightarrow cos^{-1}\left ( \frac{3}{5} \right )=2\theta\]
\[\Rightarrow \frac{3}{5}=cos2\theta \Rightarrow \frac{1-tan^{2}\theta }{1+tan^{2}\theta }=\frac{3}{5}\]
\[5-5tan^{2}\theta =3+3tan^{2}\theta\]
\[\Rightarrow 8tan^{2}\theta =2\Rightarrow \; \; tan\theta =\frac{1}{2} \Rightarrow \theta =tan^{-1}\left ( \frac{1}{2} \right ) \]
\[\Rightarrow \frac{1}{2}cos^{-1}\left ( \frac{3}{5} \right )=tan^{-1}\left ( \frac{1}{2} \right )\: \: .......\: \: (2)\]
\[Let\; \; sin^{-1}\left (\frac{1}{\sqrt{5}} \right )=\theta \Rightarrow sin\theta =\frac{1}{\sqrt{5}}\]
\[\Rightarrow tan\theta =\frac{1}{2}\Rightarrow \theta =tan^{-1}\left ( \frac{1}{2} \right )\]
\[\Rightarrow sin^{-1}\left ( \frac{1}{\sqrt{5}} \right )=tan^{-1}\left ( \frac{1}{2} \right )\: \: .....\: (3)\]
Equations (1), (2) and (3) proves the given statement

 Question 12: Prove the following

\[(i)\; \; sin\left [ cot^{-1}\left \{ cos(tan^{-1}x) \right \} \right ]=\sqrt{\frac{x^{2}+1}{x^{2}+2}}\]
\[(ii)\; \; cos\left [ tan^{-1}\left \{ sin(cot^{-1}x) \right \} \right ]=\sqrt{\frac{x^{2}+1}{x^{2}+2}}\]
Question 13: Prove the following
\[tan^{-1}\left ( \frac{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}} \right )=\frac{\pi }{4}+\frac{1}{2}cos^{-1}x^{2}\]
Question 14: Find the value of x if
\[cos^{-1}\left ( \frac{x^{2}-1}{x^{2}+1} \right )+tan^{-1}\left ( \frac{2x}{x^{2}-1} \right )=\frac{2\pi }{3}\; \; ....\; \; Ans,\;x=\sqrt{3}\]
Question 15: Find the value of the following
\[cot\frac{1}{2}\left [ cos^{-1} \frac{2x}{1+x^{2}}+sin^{-1}\frac{1-y^{2}}{1+y^{2}}\right ]\; \; .....Ans\; \frac{x+y}{1-xy}\]
Question 16: Solve for the value of x
\[tan^{-1}(x-1)+tan^{-1}x+tan^{-1}(x+1)=tan^{-1}3x\\\; \; .....\; Ans:\; x=0,\; \pm \frac{1}{2}\]
Question 17: \[If\; \theta =sin^{-1}\left \{ sin(-600^{o}) \right \},\; then\; find\; the\; value\; of\; \theta\\\; ....,Ans=\; \frac{\pi }{3}\]
Q 18. Simplifying the following
\[(i)\; \; \; cos^{-1}\left ( \frac{3}{5}cosx+\frac{4}{5}sinx \right )\; \; ....\; \; Ans\; \left [ x-tan^{-1}\frac{4}{3} \right ]\]
\[(ii)\; \; \; sin^{-1}\left ( \frac{5}{13}cosx+\frac{12}{13}sinx \right )\]
Solution (ii):\[Let\; \; \frac{5}{13}=rsin\theta \; \; and\; \; \frac{12}{13}=rcos\theta\]\[\sqrt{\left ( \frac{5}{13} \right )^{2}+\left ( \frac{12}{13} \right )^{2}}= \sqrt{r^{2}sin^{2}\theta +r^{2}cos^{2}\theta}\]\[\sqrt{\frac{25}{169}+\frac{144}{169}}=\sqrt{r^{2}\left ( sin^{2}\theta +cos^{2}\theta \right )}\] \[\sqrt{\frac{169}{169}}=r=1\]\[sin\theta =\frac{5}{13}\: \: and\: \: cos\theta =\frac{12}{13}\]\[tan\theta =\frac{sin\theta }{cos\theta }=\frac{5}{12}\Rightarrow \theta =tan^{-1}\left ( \frac{5}{12} \right )\] \[sin^{-1}\left ( \frac{5}{13}cosx+\frac{12}{13}sinx \right )=sin^{-1}\left ( sin\theta \; cosx+cos\theta\; sinx \right )\] \[=sin^{-1}\left [ sin\left ( x+\theta \right ) \right ]=x+\theta =x+tan^{-1}\frac{5}{12}\]

  Question 19: \[If\;\; (tan^{-1}x)^{2}+(cot^{-1}x)^{2}=\frac{5\pi^{2} }{8},\; then\; find\; \; x.\]

Solution : \[ (tan^{-1}x)^{2}+(cot^{-1}x)^{2}=\frac{5\pi^{2} }{8}, \]
\[(tan^{-1}x)^{2}+(cot^{-1}x)^{2}+2tan^{-1}xcot^{-1}x-2tan^{-1}xcot^{-1}x =\frac{5\pi^{2} }{8}\]
\[(tan^{-1}x+cot^{-1}x)^{2}-2tan^{-1}xcot^{-1}x =\frac{5\pi^{2} }{8}\]
\[\left (\frac{\pi }{2} \right )^{2}-2tan^{-1}x\left ( \frac{\pi }{2}-tan^{-1}x \right ) =\frac{5\pi^{2} }{8}\]
\[\frac{\pi^{2} }{4}-\pi tan^{-1}x+2(tan^{-1}x)^{2}=\frac{5\pi ^{2}}{8}\]
\[2(tan^{-1}x)^{2}-\pi tan^{-1}x-\frac{3\pi ^{2}}{8} =0\]
\[Putting\; \; tan^{-1}x = p,\; we\; get\]
\[2p^{2}-\pi p-\frac{3\pi ^{2}}{8}=0\\\\16p^{2}-8\pi p-3\pi^{2}=0\]
\[16p^{2}-12\pi p+4\pi p-3\pi ^{2}=0\\4p(4p-3\pi )+\pi (4p-3\pi )=0\]
\[(4p-3\pi )(4p+\pi)= 0\]
\[p=\frac{3\pi }{4},\: \: or\: \: p=\frac{-\pi }{4}\]
\[p=tan^{-1}x=\frac{3\pi }{4},\: \: or\: \: p=tan^{-1}x=\frac{-\pi }{4}\]
\[But\; tan^{-1}x=\frac{3\pi }{4}\; is\; rejected\; \because \frac{-\pi }{2}<x<\frac{\pi }{2}\]
\[\Rightarrow tan^{-1}x=-\frac{\pi }{4}\\ x=tan\left (-\frac{\pi }{4} \right )=-tan\left ( \frac{\pi }{4} \right )=-1\]

 


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