Inverse Trigonometric Functions Test

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Inverse Trigonometric Functions Test - 37

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Weekly Quiz Competition

Question 1
1 / -0

Find the values of $$\cos^{-1}\left(\cos \dfrac {7\pi}{6}\right)$$ is equal to

A
$$\dfrac {7\pi}{6}$$

B
$$\dfrac {5\pi}{6}$$

C
$$\dfrac {\pi}{3}$$

D
$$\dfrac {\pi}{6}$$

Solution

Let $$y=\cos^{-1}\left(\cos\dfrac{7\pi}{6}\right)$$
$$\cos y=\cos\left(\dfrac{7\pi}{6}\right)$$
$$\cos y=\cos (210^{o})$$
We know that range of principal values is $$\cos $$ is $$(0, \pi)$$
$$i.e (0^{o}, 180^{o})$$
Hence $$4=210^{o}$$ not possible
Now,
$$\cos y=\cos (210^{o})$$
$$\cos y=\cos (360-150^{o})$$
$$\cos y=\cos (150^{o})$$
$$\cos y=\cos \left(\dfrac{5\pi}{5}\right)$$

$$4=\dfrac{5\pi}{6}$$
Which is in range of principal values of $$\cos ^{-1}$$ i.e $$(0, \pi)$$
$$\therefore y=\dfrac{5\pi}{6}$$
Question 2
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The value of $$3\tan^{-1}\dfrac {1}{2}+2\tan^{-1}\dfrac {1}{5}+\sin^{-1}\dfrac {142}{65\sqrt {5}}$$ is

A
$$\dfrac {\pi}{4}$$

B
$$\dfrac {\pi}{2}$$

C
$$\pi$$

D
$$none\ of\ these$$

Solution

$$\Rightarrow 3\tan^{-1}\dfrac{1}{2}+2\tan^{-1}\dfrac{1}{5}+\sin^{-1}\dfrac{142}{65\sqrt{5}}$$
$$\Rightarrow \left(\tan^{-1}\dfrac{1}{\alpha}+\tan^{-1}\dfrac{1}{\alpha}\right)+\tan^{-1}\dfrac{1}{\alpha}+\left(\tan^{-1}\dfrac{1}{5}+\tan^{-1}\dfrac{1}{5}\right)+\tan^{-1}\left(\dfrac{142}{31}\right)$$
$$\Rightarrow \left(\tan^{-1}\dfrac{4}{3}+\tan^{-1}\dfrac{1}{\alpha}\right)+\tan^{-1}\left(\dfrac{5}{12}\right)+\tan^{-1}\left(\dfrac{142}{31}\right)$$
$$\Rightarrow \tan^{-1}\left(\dfrac{11}{2}\right)+\tan^{-1}\left(\dfrac{5}{12}\right)+\tan^{-1}\left(\dfrac{142}{31}\right)$$
$$\Rightarrow \pi +\tan^{-1}\left(\dfrac{-142}{31}\right)+\tan^{-1}\left(\dfrac{142}{31}\right)$$
$$\Rightarrow \pi$$
Important property used
$$\tan^{-1}(x)+\tan^{-1}(y)=\tan^{-1}\left(\dfrac{x+y}{1-xy}\right)$$
for $$x > o, y > o$$ $$xy < 1$$
$$\tan^{-1}(x)+\tan^{-1}(y)=\pi +\tan^{-1}\left(\dfrac{x+y}{1-xy}\right)$$
for $$x > 0, y > 0, xy > 1$$.
Question 3
1 / -0

Let in $$\Delta ABC, \, \angle A = \dfrac{\pi}{2}$$. Then value of $$\tan^{-1} \dfrac{b}{a + c} + \tan^{-1} \dfrac{c}{a + b}$$ equals

A
$$\dfrac{\pi}{2}$$

B
$$\dfrac{\pi}{4}$$

C
$$\dfrac{\pi}{3}$$

D
None of these

Solution

In $$\Delta ABC$$ ,
$$\angle A=90^{\circ}$$

So, from Pythagorus theorem,
$$a^{2}=b^{2}+c^{2}$$ .............. (1)

$$\tan^{-1}\cfrac{b}{a+c}+\tan^{-1}\cfrac{c}{a+b}$$

$$=\tan^{-1}\cfrac{\cfrac{b}{a+c}+\cfrac{c}{a+b}}{1-\cfrac{b}{a+c}\cdot \cfrac{c}{a+b}}$$

$$=\tan^{-1}\cfrac{ab+b^{2}+ac+c^{2}}{a^{2}+ab+ac}$$

$$=\tan^{-1}\cfrac{ab+ac+a^{2}}{a^{2}+ab+ac}$$ from equation (1)

$$=\tan^{-1}(1)$$

$$=\cfrac{\pi }{4}$$
Question 4
1 / -0

Find the approximate value of $$\sin^{-1}(0.51)$$ given that $$\sqrt {3}=1.7321$$

A
$$\dfrac {\pi}{6}+0.011541$$

B
$$\dfrac {\pi}{3}+0.011541$$

C
$$\dfrac {\pi}{6}+0.011547$$

D
$$\dfrac {\pi}{3}+0.011547$$

Solution

$$\sin^{-1}\left(0.5+0.01\right)=\sin^{1}\left(0.51\right)$$
$$\therefore$$ Approx value$$=f \left(x\right)+hf’\left(x\right)$$
Here $$x=0.5$$
$$h=0.01$$
$$f \left(u\right)=\sin^{1}x$$
$$f‘ \left(u\right)=\dfrac{1}{\sqrt{1-x^{2}}}$$
$$\therefore f \left(0.5\right)=\sin^{-1}\left(0.5\right)=\frac{\pi}{6}$$
$$\therefore f’ \left(0.5\right)=\dfrac{1}{\sqrt{1-\left(0.5\right)^{2}}}$$
$$=\dfrac{1}{\sqrt{0.75}}$$
$$=\boxed{\dfrac{2}{\sqrt{3}}}$$
$$\therefore$$ Approx value $$= \dfrac{\pi}{6}+ \dfrac{2}{\sqrt{3}} \times \left(\dfrac{1}{100}\right)$$
$$\dfrac{\pi}{6}+\dfrac{2}{100\sqrt{3}}$$
$$\boxed{ =\dfrac{\pi}{6}+0.011541}$$
Question 5
1 / -0

$${\sin ^{ - 1}}(\cos \left( {{{\sin }^{ - 1}}x} \right)) + {\cos ^{ - 1}}(\sin \left( {{{\cos }^{ - 1}}x} \right)){\text{is}}\;{\text{equal}}\;{\text{to}}{\text{.}}$$

A
$$\frac{\pi }{2}$$

B
$$\frac{\pi }{4}$$

C
$$\frac{{3\pi }}{4}$$

D
0

Solution

Given

$$\sin ^{-1} (\cos \sin ^{-1} (x)) +\cos ^{-1} \left(\sin \left(\cos ^{-1} x\right)\right)$$

$$ \bigg( \left(\sin ^{-1} x+\cos ^{-1} x \right) =\dfrac \pi 2\bigg)$$

$$\dfrac \pi 2- \cos ^{-1} (\cos \sin ^{-1} (x)) + \dfrac \pi 2-\sin ^{-1} \left(\sin \left(\cos ^{-1} x\right)\right)$$

$$\pi -\sin ^{-1} x-\cos ^{-1} x$$

$$\bigg( \cos ^{-1} \cos x =x ,\sin ^{-1} \sin x=x\bigg)$$

$$\pi - \left(\sin ^{-1} x+\cos ^{-1} x \right)$$

$$ \bigg( \left(\sin ^{-1} x+\cos ^{-1} x \right) =\dfrac \pi 2\bigg)$$

$$\pi-\dfrac \pi 2$$

$$\implies \dfrac \pi 2$$
Question 6
1 / -0

$$\cot ^{ -1 }{ \left( \dfrac { \sqrt { 1-\sin { x } } +\sqrt { 1+\sin { x } } }{ \sqrt { 1-\sin { x } } -\sqrt { 1+\sin { x } } } \right) }$$=....$$\left( 0<x<\dfrac { \pi }{ 2 } \right)$$

A
$$\dfrac { x }{ 2 } $$

B
$$\dfrac { \pi }{ 2 } -2x$$

C
$$2\pi -x$$

D
$$\pi -\dfrac { x }{ 2 } $$

Solution

We have,
$$\Rightarrow \cot^{-1}\left(\dfrac{\sqrt{1-\sin x}+\sqrt{1+\sin x}}{\sqrt{1-\sin x}-\sqrt{1+\sin x}}\right)$$
$$\Rightarrow \cot^{-1}\left(\dfrac{(\sqrt{1-\sin x}+\sqrt{1+\sin x})^2}{(1-\sin x)-(1+\sin x)}\right)$$
$$\Rightarrow \cot^{-1}\left(\dfrac{1-\sin x+1+\sin x+2\sqrt{1-\sin^2x}}{1-\sin x-1-\sin x}\right)$$
$$\Rightarrow \cot^{-1}\left(\dfrac{2+2\sqrt{\cos^2x}}{-2\sin x}\right)$$
$$\Rightarrow \cot^{-1}\left(\dfrac{1+\cos x}{-\sin x}\right)$$
$$\Rightarrow \cot^{-1}\left(\dfrac{1+2\cos^2 x/2-1}{-2\sin\dfrac{x}{2}\cos \dfrac{x}{2}}\right)$$
$$\Rightarrow \cot^{-1}\left(\dfrac{\cos x/2}{\sin x/2}\right)$$
$$\Rightarrow \cot^{-1}(-\cot x/2)$$
$$\Rightarrow \cot^{-1}(\cot (\pi -x/2))$$
$$\Rightarrow \pi -x/2$$.
Question 7
1 / -0

Find the value of $${\sin ^{ - 1}}\left( 1 \right)$$

A
$$\dfrac{\pi}{3}$$

B
$$\dfrac{5\pi}{2}$$

C
$$\dfrac{3\pi}{2}$$

D
$$\dfrac{\pi}{2}$$

Solution

value of$$ \sin^{-1}(1) $$
$$\sin x$$ is in vertible form
$$\left [ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right ]$$ in this range only
$$\sin \dfrac{\pi }{2}=1$$
$$\sin^{-1}(1)=\dfrac{\pi }{2}.$$
Question 8
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If $${\tan ^{ - 1}}\dfrac{1}{{a - 1}} = {\tan ^{ - 1}}\dfrac{1}{x} + {\tan ^{ - 1}}\dfrac{1}{{{a^2} - x + 1}},\,then\,x\,is\,$$

A
$$a$$

B
$${a^{3\,}}$$

C
$${a^2} - a + 1\,$$

D
$${a^2} + a - 1$$

Solution

Given,
$${\tan ^{ - 1}}\dfrac{1}{{a - 1}} = {\tan ^{ - 1}}\dfrac{1}{x} + {\tan ^{ - 1}}\dfrac{1}{{{a^2} - x + 1}}$$

or, $${\tan ^{ - 1}}\dfrac{1}{{a - 1}} = {\tan ^{ - 1}}\dfrac{\dfrac{1}{x}+\dfrac{1}{a^2-x+1}}{1-\dfrac{1}{x(a^2-x+1)}} $$

or, $${\tan ^{ - 1}}\dfrac{1}{{a - 1}} = {\tan ^{ - 1}}\dfrac{a^2+1}{a^2x-x^2+x-1} $$

or, $$(a-1)(a^2+1)=a^2 x-x^2+x-1$$

or, $$a^3+a-a^2-1=a^2 x-x^2+x-1$$

or, $$a^2(a-x)-(a^2-x^2)+(a-x)=0$$

or, $$(a-x)(a^2-a-x+1)=0$$

or, $$x=a, a^2-a+1$$.
Question 9
1 / -0

$${\cot ^{ - 1}}\left( {2 + \sqrt 3 } \right) = $$

A
$$\frac{\pi }{{12}}$$

B
$$\frac{\pi }{{15}}$$

C
$$\frac{\pi }{5}$$

D
$$\frac{{3\pi }}{{10}}$$

Solution

Solution :- given
$$ \displaystyle \Rightarrow cot^{-1}(2+\sqrt{3}) $$

$$ \displaystyle \Rightarrow cot^{-1}(\frac{2(2+\sqrt{3})}{2}) $$ multiply num and den by 2

$$ \displaystyle \Rightarrow cot^{-1}(\frac{4+2\sqrt{3}}{3-1}) $$

$$ \displaystyle \Rightarrow cot^{-1}(\frac{3+1+2\sqrt{3}}{(\sqrt{3})^{2}-(1)^{2}}) $$

$$ \displaystyle \Rightarrow cot^{-1}(\frac{(\sqrt{3})^{2}+(1)^{2}+2.1.\sqrt{3}}{(\sqrt{3}+1)(\sqrt{3}-1)}) $$

$$ \displaystyle \Rightarrow cot^{-1}(\frac{(\sqrt{3}+1)^{2}}{(\sqrt{3}+1)(\sqrt{3}-1)}) $$

$$ \displaystyle \Rightarrow cot^{-1}(\frac{\sqrt{3}+1}{\sqrt{3}-1}) $$

$$ \displaystyle \Rightarrow cot^{-1}(\frac{(cot45\,cot30+1)}{cot30-cot45}) $$

$$ \displaystyle = cot^{-1} (cot (45-30))$$

$$= 45^{\circ}-30^{\circ} $$

$$ \displaystyle = 15^{\circ} $$

$$ \displaystyle = \frac{\pi }{12} $$ Ans
Question 10
1 / -0

$$cot^{-1}(\sqrt{cos\alpha })-tan^{-1}(\sqrt{cos\alpha })=x$$ , then sin x is equal to

A
$$\tan^{2}\dfrac{\alpha }{2}$$

B
$$\cot^{2}\dfrac{\alpha }{2}$$

C
$$\tan \alpha $$

D
$$\cot\dfrac{\alpha }{2}$$

Solution

$$\cot^{-1}(\sqrt{\cos \alpha})-\tan^{-1}(\sqrt{\cos \alpha})=x$$
$$\dfrac{\pi}{2}-\tan^{-1}(\sqrt{\cos \alpha})-\tan^{-1}(\sqrt{\cos \alpha})=x$$ (because $$\cot^{-1}x+\tan^{-1}x=\dfrac{\pi}{2}$$)
$$\Rightarrow \dfrac{\pi}{2}-2\tan^{-1}(\sqrt{\cos \alpha})=x$$
$$\Rightarrow \sin x=\sin\left(\dfrac{\pi}{2}-2\tan^{-1}(\sqrt{\cos\alpha})\right)$$
Let's say $$\tan^{-1}(\sqrt{\cos \alpha})=\theta$$
$$\Rightarrow \tan\theta =\sqrt{\cos\alpha}$$
$$\sin x=sin\left(\dfrac{\pi}{2}-2\theta\right)=\cos 2\theta$$
$$=\dfrac{1-\tan^2\theta}{1+\tan^2\theta}$$
$$=\dfrac{1-\cos\alpha}{1+\cos\alpha}$$
$$\sin x=\dfrac{1-\cos \alpha}{1+\cos \alpha}$$
$$=\dfrac{2\sin^2\alpha/2}{2\cos^2\alpha/2}$$
$$\Rightarrow \sin x=\tan^2\dfrac{\alpha}{2}$$.

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Inverse Trigonometric Functions Test - 37

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Inverse Trigonometric Functions Test - 37

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SHARING IS CARING

Question 1

$$\dfrac {7\pi}{6}$$

$$\dfrac {5\pi}{6}$$

$$\dfrac {\pi}{3}$$

$$\dfrac {\pi}{6}$$

Solution

Question 2

$$\dfrac {\pi}{4}$$

$$\dfrac {\pi}{2}$$

$$\pi$$

$$none\ of\ these$$

Solution

Question 3

$$\dfrac{\pi}{2}$$

$$\dfrac{\pi}{4}$$

$$\dfrac{\pi}{3}$$

None of these

Solution

Question 4

$$\dfrac {\pi}{6}+0.011541$$

$$\dfrac {\pi}{3}+0.011541$$

$$\dfrac {\pi}{6}+0.011547$$

$$\dfrac {\pi}{3}+0.011547$$

Solution

Question 5

$$\frac{\pi }{2}$$

$$\frac{\pi }{4}$$

$$\frac{{3\pi }}{4}$$

0

Solution

Question 6

$$\dfrac { x }{ 2 } $$

$$\dfrac { \pi }{ 2 } -2x$$

$$2\pi -x$$

$$\pi -\dfrac { x }{ 2 } $$

Solution

Question 7

$$\dfrac{\pi}{3}$$

$$\dfrac{5\pi}{2}$$

$$\dfrac{3\pi}{2}$$

$$\dfrac{\pi}{2}$$

Solution

Question 8

$$a$$

$${a^{3\,}}$$

$${a^2} - a + 1\,$$

$${a^2} + a - 1$$

Solution

Question 9

$$\frac{\pi }{{12}}$$

$$\frac{\pi }{{15}}$$

$$\frac{\pi }{5}$$

$$\frac{{3\pi }}{{10}}$$

Solution

Question 10

$$\tan^{2}\dfrac{\alpha }{2}$$

$$\cot^{2}\dfrac{\alpha }{2}$$

$$\tan \alpha $$

$$\cot\dfrac{\alpha }{2}$$

Solution

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