### Mathematics Assignments | PDF | 8 to 12

PDF Files of Mathematics Assignments From VIII Standard to XII Standard PDF of mathematics Assignments for the students from VIII standard to XII standard.These assignments are strictly according to the CBSE and DAV Board Final question Papers

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

Maths Assignment
Inverse Trigonometric Functions

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

Question 1: Find the Principal Value of the following functions

$\bg{black}\bg{black}\mathbf{ (i)\:\: \: \: \: cos^{-1}\left ( cos\frac{7\pi }{6} \right )}$   Ans: 5Ï€/6
$\bg{black}\mathbf{(ii)\:\: \: \: \: cos^{-1}\left ( cos\frac{2\pi }{3} \right )+sin^{-1}\left ( sin\frac{2\pi }{3} \right )$
Ans : Ï€
$\bg{black}\mathbf{(iii)\:\: \: \: \: tan^{-1}\left ( tan\frac{2\pi }{3} \right )+tan^{-1}\left ( tan\frac{9\pi }{8} \right )}$
Ans: -5Ï€/24
$\bg{black}\mathbf{(iv)\:\: \: \: \: sec^{-1}\left ( 2sin\frac{3\pi }{4} \right )}$
Ans: Ï€/4
Question 2:
(i)  Find the value of :
$\bg{black}\mathbf{ 4[cot^{-1}(3)+cosec^{-1}(\sqrt{5})]}$
Ans : Ï€
(ii) Find the value of:   $\large \bg{black}\mathbf{sin\left ( 2sin^{-1}\frac{3}{5} \right )}$
Question 3: Find the value of the following
$\large \bg{black}\mathbf{2sin^{-1}\frac{1}{2}+3tan^{-1}(-1)+2cos^{-1}\left ( \frac{-1}{2} \right )+4sec^{-1}(\sqrt{2}) }$
Ans: 23Ï€/12
Question 4: Find the value of the following
$\large \bg{black}\mathbf{4\left [ 2sin^{-1}\left ( \frac{-1}{2} \right )+5tan^{-1}(1)-3cos^{-1}\left ( \frac{1}{2} \right ) \right ]+\frac{1}{2}cos^{-1}\left ( \frac{-\sqrt{3}}{2} \right )}$
Ans: Ï€/12
Question 5: Find the value of the following
$\bg{black}\mathbf{4cos^{-1}\left ( \frac{1}{2} \right )+2sin^{-1}\left ( \frac{1}{2} \right )+3tan^{-1}(-1)+2cos^{-1}\left ( \frac{\sqrt{3}}{2} \right )}$
Ans: 5Ï€/4
Question 6: Find the value of the following

$\bg{black}\mathbf{sin^{-1}\left ( sin\frac{3\pi }{5} \right )+sin^{-1}\left ( sin\frac{4\pi }{5} \right )}$
Ans: 3
Ï€/5
Question 7:
Find the domain of sin-1(x2 - 4) . Also find its range
Solution Hint
Domain of sin-1 x is [-1, 1], so for the given function we have
-1≤ x2 - 4 ≤ 1

3≤ x2 ≤ 5

Taking square root we get

$\bg{black}\mathbf{\pm\sqrt{3}\leq x\leq\pm\sqrt{5}}$.  now we have

$\bg{black}\mathbf{x\leq\pm\sqrt{5}\Rightarrow-\sqrt{5}\leq x\leq\sqrt{5}}$

$\bg{black}\mathbf{x\geq\pm\sqrt{3}\Rightarrow x\leq-\sqrt{3}\;\:and\:\:x\geq\sqrt{3}}$

On the number line common region of these inequilities is shown with the red line

So the domain of the given function is given by

$\bg{black}\mathbf{[-\sqrt{5},-\sqrt{3}]\cup[\sqrt{3},\sqrt{5}]}$
Range = [-Ï€/2, Ï€/2]
Question 8
Evaluate: $\bg{black}\mathbf{sec^{2}\left(tan^{-1}\frac{1}{2}\right)+cosec^{2}\left(cot^{-1}\frac{1}{3}\right)}$
Solution: $\bg{black}\mathbf{sec^{2}\left(tan^{-1}\frac{1}{2}\right)+cosec^{2}\left(cot^{-1}\frac{1}{3}\right)}$

= $\bg{black}\mathbf{\left[1+tan^{2}\left(tan^{-1}\frac{1}{2}\right)\right]+\left[1+cot^{2}\left(cot^{-1}\frac{1}{3}\right)\right]}$
= $\bg{black}\mathbf{\left[1+\left(\frac{1}{2}\right)^{2}\right]+\left[1+\left(\frac{1}{3}\right)^{2}\right]}$
= 85/36
Question 9
Express   $\bg{black}\mathbf{tan^{-1}\left(\frac{cosx}{1-sinx}\right),where\;-\frac{\pi}{2} in the simplest form.
Solution:
$\bg{black}\mathbf{y=\left[tan^{-1}\left(\frac{cosx}{1-sinx}\right)\right]}$

## $\bg{black}\mathbf{y=\frac{\pi}{4}+\frac{x}{2}}$

Question 10:  Find the value of k if
$\bg{black}\mathbf{sin^{-1}\left[k\:tan\left(2cos^{-1}\frac{\sqrt{3}}{2}\right)\right]=\frac{\pi}{3}}$
Ans: k = 1/2
Question 11: Find the value of   $\large \bg{black}\mathbf{sin^{-1}\left[cos\left(\frac{33\pi}{5}\right)\right]}$
Solution Hint:
$\large \bg{black}\mathbf{sin^{-1}\left[cos\left(\frac{33\pi}{5}\right)\right]=sin^{-1}\left[cos\left(6\pi+\frac{3\pi}{5}\right)\right]}$ $\large \bg{black}\mathbf{sin^{-1}\left[cos\left(\frac{33\pi}{5}\right)\right]=sin^{-1}\left[cos\left(\frac{3\pi}{5}\right)\right]}$ $\large \bg{black}\mathbf{sin^{-1}\left[cos\left(\frac{33\pi}{5}\right)\right]=sin^{-1}\left[sin\left(\frac{\pi}{2}-\frac{3\pi}{5}\right)\right]}$ $\large \bg{black}\mathbf{sin^{-1}\left[cos\left(\frac{33\pi}{5}\right)\right]=\frac{\pi}{2}-\frac{3\pi}{5}}=\frac{-\pi}{10}$

Question 12: Prove the following
$\large \bg{black}\mathbf{tan^{-1}\left ( \frac{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}} \right )=\frac{\pi }{4}+\frac{1}{2}cos^{-1}x^{2}}$
Question 13: Prove the following
$\large \bg{black}\mathbf{(i)\; \; sin\left [ cot^{-1}\left \{ cos(tan^{-1}x) \right \} \right ]=\sqrt{\frac{x^{2}+1}{x^{2}+2}}}$ $\large \bg{black}\mathbf{(ii)\; \; cos\left [ tan^{-1}\left \{ sin(cot^{-1}x) \right \} \right ]=\sqrt{\frac{x^{2}+1}{x^{2}+2}}}$
Question 14: Find the value of x if
$\large \bg{black}\mathbf{cos^{-1}\left ( \frac{x^{2}-1}{x^{2}+1} \right )+tan^{-1}\left ( \frac{2x}{x^{2}-1} \right )=\frac{2\pi }{3}}$
Question 15: Find the value of the following

$\large \bg{black}\mathbf{cot\frac{1}{2}\left [ cos^{-1} \frac{2x}{1+x^{2}}+sin^{-1}\frac{1-y^{2}}{1+y^{2}}\right ]}$

$\large \bg{black}\mathbf{Ans\; \frac{x+y}{1-xy}}$
Question 16: Solve for the value of x
$\large \bg{black}\mathbf{tan^{-1}(x-1)+tan^{-1}x+tan^{-1}(x+1)=tan^{-1}3x}$
$\large \bg{black}\mathbf{Ans:\; x=0,\; \pm \frac{1}{2}}$

## Extra Questions for more Practice

Question 1: Write the following functions in the simplest form

$(i)\; \; \; tan^{-1}\left ( \sqrt{\frac{a-x}{a+x}} \right )\; \; ....\; \; Ans:\; \left ( \frac{1}{2}cos^{-1}\frac{x}{a} \right )$ $(ii)\; \; \; cos^{-1}\left ( \frac{x}{\sqrt{x^{2}+a^{2}}} \right )\; \; .....\; \; Ans:\; cot^{-1}\left (\frac{x}{a} \right )$ $(iii)\; \; \; sin^{-1}\left ( \frac{x}{\sqrt{x^{2}+a^{2}}} \right )\; \; ........\; \; Ans:\; tan^{-1}\frac{x}{a}$
Question 2: Simplify the followings

$(i)\; \; \; sin^{-1}\left ( \frac{sinx+cosx}{\sqrt{2}} \right ) \; ........\; \; Ans:\; x+\frac{\pi }{4}$
$\begin{matrix}(ii)\; \; \; cos^{-1}\left ( \frac{sinx+cosx}{\sqrt{2}} \right ) \; ........\; \; Ans:\; x-\frac{\pi }{4}\end{matrix}$
Question 4: Find the value of x if

$2tan^{-1}(sinx)=tan^{-1}(2secx)\; \; .......\; \; Ans:\; \; x=\frac{\pi }{4}$
Question 5: 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 6: 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 7:

$\theta =sin^{-1}\left \{ sin(-600^{o}) \right \}$ then find the value of Î¸

$Ans=\; \frac{\pi }{3}$

Question 8. 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 9:

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

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}

$\Rightarrow tan^{-1}x=-\frac{\pi }{4}$

$x=tan\left (-\frac{\pi }{4} \right )=-tan\left ( \frac{\pi }{4} \right )=-1$