Inverse Trigonometric Functions Test

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Inverse Trigonometric Functions Test - 38

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Weekly Quiz Competition

Question 1
1 / -0

The value of $${\sin ^{ - 1}}\left( {\cos \dfrac{{33\pi }}{5}} \right)$$ is

A
$$\dfrac{{3\pi }}{5}$$

B
$$\dfrac{{7\pi }}{5}$$

C
$$\dfrac{\pi }{{10}}$$

D
$$ - \dfrac{\pi }{{10}}$$

Solution

$$\sin^{-1}\left(\cos\dfrac{33\pi}{5}\right)=\sin^{-1}\left(\cos\left(6\pi+\dfrac{3\pi}{5}\right)\right)$$

$$=\sin^{-1}\left(\cos\left(\dfrac{3\pi}{5}\right)\right)$$

$$\sin^{-1}\left(\cos\left(\dfrac{\pi}{2}+\dfrac{\pi}{10}\right)\right)$$

$$=\sin^{-1}\left(-\sin\left(\dfrac{\pi}{10}\right)\right)$$

$$\therefore \sin^{-1}\left(\cos\dfrac{33\pi}{5}\right)=\dfrac{-\pi}{10}$$
Question 2
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If minimum value of $${\left( {{{\sin }^{ - 1}}x} \right)^2} + {\left( {{{\cos }^{ - 1}}x} \right)^2} = \frac{{{\pi ^2}}}{K}$$ then the value of $$K$$ is

A
$$4$$

B
$$8$$

C
$$6$$

D
$$16$$

Solution
Question 3
1 / -0

The value of $$\cos(\tan^{-1}(\tan2))$$

A
$$\dfrac{1}{\sqrt{5}}$$

B
$$-\dfrac{1}{\sqrt{5}}$$

C
$$\cos 2$$

D
$$-\cos 2$$

Solution

$$=\cos ( \tan^{-1} (\tan 2 ) ) = \cos 2$$
Question 4
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If $$\sin ^{ -1 }{ x } =\cot ^{ -1 }{ x } $$ then

A
$${ x }^{ 2 }=\cfrac { \sqrt { 5 } -1 }{ 2 } $$

B
$${ x }^{ 2 }=\cfrac { \sqrt { 5 } +1 }{ 2 } $$

C
$${ x }^{ }=\cfrac { \sqrt { 5 } +1 }{ 2 } $$

D
$${ x }^{ }=\cfrac { \sqrt { 5 } -1 }{ 2 } $$

Solution

$$\mathbf{{\text{Step - 1: Writing co}}{{\text{t}}^{{\text{ - 1}}}}{\text{x in terms of si}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}}$$
$${\text{Let co}}{{\text{t}}^{{\text{ - 1}}}}{\text{x = }}\theta $$
  $$ \Rightarrow {\text{ x = cot }}\theta $$
  $${\text{We know that, 1 + co}}{{\text{t}}^{\text{2}}}\theta {\text{ = cose}}{{\text{c}}^{\text{2}}}\theta $$
  $$\therefore {\text{ 1 + }}{{\text{x}}^{\text{2}}}{\text{ = }}\dfrac{1}{{{\text{si}}{{\text{n}}^{\text{2}}}\theta }}$$
  $$ \Rightarrow {\text{ si}}{{\text{n}}^{\text{2}}}\theta {\text{ = }}\dfrac{{\text{1}}}{{{\text{1 + }}{{\text{x}}^{\text{2}}}}}$$
  $$ \Rightarrow {\text{ sin }}\theta {\text{ = }}\sqrt {\dfrac{{\text{1}}}{{{\text{1 + }}{{\text{x}}^{\text{2}}}}}} $$
  $$ \Rightarrow {\text{ }}\theta {\text{ = si}}{{\text{n}}^{{\text{ - 1}}}}\left( {\sqrt {\dfrac{{\text{1}}}{{{\text{1 + }}{{\text{x}}^{\text{2}}}}}} } \right)$$
  $$ \Rightarrow {\text{ co}}{{\text{t}}^{{\text{ - 1}}}}{\text{x = si}}{{\text{n}}^{{\text{ - 1}}}}\left( {\sqrt {\dfrac{{\text{1}}}{{{\text{1 + }}{{\text{x}}^{\text{2}}}}}} } \right)$$
$$\mathbf{{\text{Step - 2: Equating it to si}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}}$$
  $${\text{si}}{{\text{n}}^{{\text{ - 1}}}}{\text{x = si}}{{\text{n}}^{{\text{ - 1}}}}\left( {\sqrt {\dfrac{{\text{1}}}{{{\text{1 + }}{{\text{x}}^{\text{2}}}}}} } \right)$$
  $$ \Rightarrow {\text{ x = }}\left( {\sqrt {\dfrac{{\text{1}}}{{{\text{1 + }}{{\text{x}}^{\text{2}}}}}} } \right)$$
  $$ \Rightarrow {\text{ }}{{\text{x}}^{\text{2}}}{\text{ = }}\dfrac{{\text{1}}}{{{\text{1 + }}{{\text{x}}^{\text{2}}}}}$$
  $$ \Rightarrow {\text{ }}{{\text{x}}^{\text{4}}}{\text{ + }}{{\text{x}}^{\text{2}}}{\text{ - 1 = 0}}$$
  $$ \Rightarrow {\text{ }}{{\text{x}}^{\text{2}}}{\text{ = }}\dfrac{{{\text{ - 1 }} \pm {\text{ }}\sqrt {{\text{1 + 4}}} }}{{\text{2}}}$$
  $$\because {\text{ }}{{\text{x}}^{\text{2}}}{\text{ cannot be negative}}$$
  $$ \Rightarrow {\text{ }}{{\text{x}}^{\text{2}}}{\text{ = }}\dfrac{{\sqrt {\text{5}} {\text{ - 1}}}}{{\text{2}}}$$
$$\mathbf{{\text{Hence, If si}}{{\text{n}}^{{\text{ - 1}}}}{\text{x = co}}{{\text{t}}^{{\text{ - 1}}}}{\text{x then }}{{\text{x}}^{\text{2}}}{\text{ = }}\dfrac{{\sqrt {\text{5}} {\text{ - 1}}}}{{\text{2}}}}$$
Question 5
1 / -0

$${\cos ^{ - 1}}\dfrac{3}{5} - {\sin ^{ - 1}}\dfrac{4}{5} = {\cos ^{ - 1}}x$$ then $$x$$ is equal to:

A
$$0$$

B
$$1$$

C
$$-1$$

D
none of these

Solution

$$\cos^{-1} \dfrac{3}{5} - \sin^{-1} \dfrac{4}{5} = \cos^{-1} x$$

$$\Rightarrow \cos^{-1} \dfrac{3}{5}- \cos^{-1} \dfrac{3}{5} = \cos^{-1} x$$

$$[\because \sin^{-1} \dfrac{4}{5} \Rightarrow \cos \theta = \dfrac{3}{5}]$$

$$\Rightarrow 0 = \cos ^{-1} x$$

$$\Rightarrow \cos 0 =x$$

$$\Rightarrow 1= x$$
Question 6
1 / -0

$$Sin^{-1}(2x(1-x^2)) = 2sin^{-1}x$$ is true if.

A
$$x\in [0,1]$$

B
[$$-\dfrac{1}{2} ,\dfrac{1}{2}$$]

C
[$$\dfrac{1}{2},- \dfrac{1}{2}$$]

D
[$$\dfrac{3}{2}, -\dfrac{3}{2}$$]

Solution
Question 7
1 / -0

The principal value of $$tan^{-1}[cot\dfrac{3\pi}{4}]$$ is :

A
$$\dfrac{-3\pi}{4}$$

B
$$\dfrac{3\pi}{4}$$

C
$$\dfrac{-\pi}{4}$$

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

Solution
Question 8
1 / -0

The greatest and least value of $${\left( {{{\sin }^{ - 1}}x} \right)^2} + {\left( {{{\cos }^{ - 1}}x} \right)^2}$$ are respectively

A
$$\dfrac{{{\pi ^2}}}{4}and0$$

B
$$\dfrac{\pi }{2}and\, - \dfrac{\pi }{2}$$

C
$$\dfrac{{5{\pi ^2}}}{4}and\dfrac{{{\pi ^2}}}{8}$$

D
$$\dfrac{{{\pi ^2}}}{4}\,and\,\dfrac{{ - \pi }}{4}$$

Solution

$$\begin{aligned} \text { let } & \sin ^ { - 1 } x = 0 \\ & = x = \sin \theta \end{aligned}$$
$$\cos ^ { - 1 } x$$ = $$\pi/2 - {\theta}$$
thus $$\left( \sin ^ { - 1 } x \right) ^ { 2 } + \left( \cos ^ { - 1 } x \right) ^ { 2 }$$
$$o ^ { 2 } + \left( \frac { \pi } { 2 } - 0 \right) ^ { 2 }$$
thus max value is = $$\frac { 5 \pi ^ { 2 } } { 4 }$$
min value is =$$\frac { \pi ^ { 2 } } { 8 }$$
Question 9
1 / -0

The number of elements in the range of
$$f(x)=\sin ^{ -1 }{ x } +\cos ^{ -1 }{ x } +\sec ^{ -1 }{ x } $$ is

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution

$$y=\sin^{-1}x+\cos^{-1}x+\sec^{-1}x$$
$$\sin^{-1}x+\cos^{-1}x\rightarrow x+(-1, 1)$$
$$\sec^{-1}x\rightarrow x\in(-\infty, -1)\cup (1, \infty)$$
$$\therefore x=-1, 1$$
$$y(-1)=\dfrac{\pi}{2}+\sec^{-1}(-1)=\dfrac{\pi}{2}+\pi =\dfrac{3\pi}{2}$$
$$y(1)=\dfrac{\pi}{2}+\sec^{-1}1=\dfrac{\pi}{2}+0=\dfrac{\pi}{2}$$
Hence, range contains two elements.
Question 10
1 / -0

The algebraic expression for $$\tan \left(\sin^{-1}\cos\ \tan^{-1}\dfrac {x}{2}\right)$$ is

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

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

C
$$\dfrac {1}{x}$$

D
$$\dfrac {2}{|x|}$$

Solution

$$\tan { \left( \sin ^{ -1 }{ \cos } \tan ^{ -1 }{ \dfrac { x }{ 2 } } \right) } $$
$$\because \tan^{-1}\dfrac{x}{2}=\cos^{-1}\dfrac{2}{\sqrt{4+x^2}}$$
$$=\tan \left( \sin^{-1}\cos \left( cos^{-1}\dfrac{2}{\sqrt{4+x^2}}\right)\right)$$
$$=\tan \left( \sin^{-1} \left( \dfrac{2}{\sqrt{4+x^2}}\right) \right)$$
$$\because \sin^{-1}\dfrac{2}{\sqrt{4+x^2}}=\tan^{-1}\dfrac{x}{2}$$
$$=\tan \left( \tan^{-1}\left( \dfrac{2}{x}\right)\right)$$
$$=\dfrac{2}{x}.$$
Hence, the answer is $$\dfrac{2}{x}.$$