Dictionary Rank of a Word | Permutations & Combinations

 PERMUTATIONS & COMBINATIONS Rank of the word or Dictionary order of the English words like COMPUTER, COLLEGE, SUCCESS, SOCCER, RAIN, FATHER, etc. Dictionary Rank of a Word Method of finding the Rank (Dictionary Order) of the word  “R A I N” Given word: R A I N Total letters = 4 Letters in alphabetical order: A, I, N, R No. of words formed starting with A = 3! = 6 No. of words formed starting with I = 3! = 6 No. of words formed starting with N = 3! = 6 After N there is R which is required R ----- Required A ---- Required I ---- Required N ---- Required RAIN ----- 1 word   RANK OF THE WORD “R A I N” A….. = 3! = 6 I……. = 3! = 6 N….. = 3! = 6 R…A…I…N = 1 word 6 6 6 1 TOTAL 19 Rank of “R A I N” is 19 Method of finding the Rank (Dictionary Order) of the word  “F A T H E R” Given word is :  "F A T H E R" In alphabetical order: A, E, F, H, R, T Words beginni

Math Assignment | Class XII | Ch-2 : Inverse Trigonometric Functions

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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