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

Basic concepts and formulas

Question 1: Find the Principal Value of the following functions

$\mathbf{ (i)\:\: \: \: \: cos^{-1}\left ( cos\frac{7\pi }{6} \right )}$ Ans: 5π/6

$\mathbf{(ii)\:\: \: \: \: cos^{-1}\left ( cos\frac{2\pi }{3} \right )+sin^{-1}\left ( sin\frac{2\pi }{3} \right )$

Ans : π

$\mathbf{(iii)\:\: \: \: \: tan^{-1}\left ( tan\frac{2\pi }{3} \right )+tan^{-1}\left ( tan\frac{9\pi }{8} \right )}$

Ans: -5π/24

$\mathbf{(iv)\:\: \: \: \: sec^{-1}\left ( 2sin\frac{3\pi }{4} \right )}$

Ans: π/4

Question 2:
(i) Find the value of : $\mathbf{ 4[cot^{-1}(3)+cosec^{-1}(\sqrt{5})]}$

Ans : π

(ii) Find the value of: $\mathbf{sin\left ( 2sin^{-1}\frac{3}{5} \right )}$

Answer : 24/25

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

Question 5: Find the value of the following
$\mathbf{4cos^{-1}\left ( \frac{1}{2} \right )+2sin^{-1}\left ( \frac{1}{2} \right )+3tan^{-1}(-1)+2cos^{-1}\left ( \frac{\sqrt{3}}{2} \right )}$
Ans: 5π/4
Question 6: Find the value of the following

$\mathbf{sin^{-1}\left ( sin\frac{3\pi }{5} \right )+sin^{-1}\left ( sin\frac{4\pi }{5} \right )}$
Ans: 3π/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

Adding 4 we get

3≤ x² ≤ 5

Taking square root we get

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

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

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

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

$\mathbf{[-\sqrt{5},-\sqrt{3}]\cup[\sqrt{3},\sqrt{5}]}$

Range = [-π/2, π/2]

Question 8
Evaluate: $\mathbf{sec^{2}\left(tan^{-1}\frac{1}{2}\right)+cosec^{2}\left(cot^{-1}\frac{1}{3}\right)}$
Solution: $\mathbf{sec^{2}\left(tan^{-1}\frac{1}{2}\right)+cosec^{2}\left(cot^{-1}\frac{1}{3}\right)}$

= $\mathbf{\left[1+tan^{2}\left(tan^{-1}\frac{1}{2}\right)\right]+\left[1+cot^{2}\left(cot^{-1}\frac{1}{3}\right)\right]}$
= $\mathbf{\left[1+\left(\frac{1}{2}\right)^{2}\right]+\left[1+\left(\frac{1}{3}\right)^{2}\right]}$
= 85/36
Question 9

Express $\mathbf{tan^{-1}\left(\frac{cosx}{1-sinx}\right),where\;-\frac{\pi}{2}<x<\frac{\pi}{2}},$ in the simplest form.

Solution:

$\mathbf{y=\left[tan^{-1}\left(\frac{cosx}{1-sinx}\right)\right]}$

$\mathbf{y=\left[tan^{-1}\left(\frac{cos^{2}x/2-sin^{2}x/2}{\left(cosx/2-sinx/2\right)^{2}}\right)\right]}$

$\mathbf{y=tan^{-1}\left[\frac{1+tanx/2}{1-tanx/2}\right]}$

$\mathbf{y=tan^{-1}\left[tan\left(\frac{\pi}{4}+\frac{x}{2}\right)\right]}$

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

Question 10: Find the value of k if
$\mathbf{sin^{-1}\left[k\:tan\left(2cos^{-1}\frac{\sqrt{3}}{2}\right)\right]=\frac{\pi}{3}}$
Ans: k = 1/2

Question 11: Find the value of $\mathbf{sin^{-1}\left[cos\left(\frac{33\pi}{5}\right)\right]}$

Answer : -π/10

Solution Hint:

$\mathbf{sin^{-1}\left[cos\left(\frac{33\pi}{5}\right)\right]=sin^{-1}\left[cos\left(6\pi+\frac{3\pi}{5}\right)\right]}$ $\mathbf{sin^{-1}\left[cos\left(\frac{33\pi}{5}\right)\right]=sin^{-1}\left[cos\left(\frac{3\pi}{5}\right)\right]}$ $\mathbf{sin^{-1}\left[cos\left(\frac{33\pi}{5}\right)\right]=sin^{-1}\left[sin\left(\frac{\pi}{2}-\frac{3\pi}{5}\right)\right]}$ $\mathbf{sin^{-1}\left[cos\left(\frac{33\pi}{5}\right)\right]=\frac{\pi}{2}-\frac{3\pi}{5}}=\frac{-\pi}{10}$

Question 12: Prove the following

$\mathbf{tan^{-1}\left ( \frac{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}} \right )=\frac{\pi }{4}+\frac{1}{2}cos^{-1}x^{2}}$

Question 13: Prove the following
$\mathbf{(i)\; \; sin\left [ cot^{-1}\left \{ cos(tan^{-1}x) \right \} \right ]=\sqrt{\frac{x^{2}+1}{x^{2}+2}}}$ $\mathbf{(ii)\; \; cos\left [ tan^{-1}\left \{ sin(cot^{-1}x) \right \} \right ]=\sqrt{\frac{x^{2}+1}{x^{2}+2}}}$

Question 14: Find the value of x if
$\mathbf{cos^{-1}\left ( \frac{x^{2}-1}{x^{2}+1} \right )+tan^{-1}\left ( \frac{2x}{x^{2}-1} \right )=\frac{2\pi }{3}}$
Answer: $\mathbf{x=\frac{\sqrt{3}-1}{\sqrt{3}+1}}$
Question 15: Find the value of the following

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

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

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Deleted questions from CBSE Syllabus

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 θ

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

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 :
$(tan^{-1}x)^{2}+(cot^{-1}x)^{2}=\frac{5\pi^{2} }{8},$

$(tan^{-1}x)^{2}+(cot^{-1}x)^{2}+2tan^{-1}xcot^{-1}x-2tan^{-1}xcot^{-1}x =\frac{5\pi^{2} }{8}$

$(tan^{-1}x+cot^{-1}x)^{2}-2tan^{-1}xcot^{-1}x =\frac{5\pi^{2} }{8}$

$\left (\frac{\pi }{2} \right )^{2}-2tan^{-1}x\left ( \frac{\pi }{2}-tan^{-1}x \right ) =\frac{5\pi^{2} }{8}$

$\frac{\pi^{2} }{4}-\pi tan^{-1}x+2(tan^{-1}x)^{2}=\frac{5\pi ^{2}}{8}$

$2(tan^{-1}x)^{2}-\pi tan^{-1}x-\frac{3\pi ^{2}}{8} =0$

$Putting\; \; tan^{-1}x = p,\; we\; get$

$2p^{2}-\pi p-\frac{3\pi ^{2}}{8}=0$