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

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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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