Inverse Trigonometric Functions Test

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Inverse Trigonometric Functions Test - 35

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Weekly Quiz Competition

Question 1
1 / -0

$$\tan \left[\dfrac{\pi}{4} + \dfrac{1}{2} \cos^{-1} \left(\dfrac{5}{7} \right) \right] + \cot \left[\dfrac{\pi}{4} + \dfrac{1}{2} \cos^{-1} \left(\dfrac{5}{7}\right) \right]$$ is equal to

A
$$\dfrac{5}{7}$$

B
$$\dfrac{10}{7}$$

C
$$\dfrac{14}{5}$$

D
$$\dfrac{7}{5}$$

Solution

$$\tan\left [ \dfrac{\pi}{4}+\dfrac{1}{2}\cos^{-1}\left ( \dfrac{5}{7} \right ) \right ]+\cot \left [ \dfrac{\pi}{4}+\dfrac{1}{2}\cos^{-1}\left ( \dfrac{5}{7} \right ) \right ] $$

$$=\tan\left [ \dfrac{\pi}{4}+\dfrac{1}{2}\cos^{-1}\left ( \dfrac{5}{7} \right ) \right ]+\tan \left [ \dfrac{\pi}{4}-\dfrac{1}{2}\cos^{-1}\left ( \dfrac{5}{7} \right ) \right ] $$

We are using the formula:
$$\tan \left [ \dfrac{\pi}{4}+\dfrac{x}{2} \right ]+\tan \left [ \dfrac{\pi}{4}-\dfrac{x}{2} \right ]=\dfrac{2}{\cos x}, x=\cos^{-1}\dfrac{a}{b}$$

$$\tan \left [ \dfrac{\pi}{4}+\dfrac{x}{2} \right ]+\tan \left [ \dfrac{\pi}{4}-\dfrac{x}{2} \right ]=\dfrac{2b}{a}$$

$$\therefore \tan\left [ \dfrac{\pi}{4}+\dfrac{1}{2}\cos^{-1}\left ( \dfrac{5}{7} \right ) \right ]+\cot \left [ \dfrac{\pi}{4}+\dfrac{1}{2}\cos^{-1}\left ( \dfrac{5}{7} \right ) \right ] =\dfrac{2\times 7}{5}$$

$$\Rightarrow \therefore \tan\left [ \dfrac{\pi}{4}+\dfrac{1}{2}\cos^{-1}\left ( \dfrac{5}{7} \right ) \right ]+\cot \left [ \dfrac{\pi}{4}+\dfrac{1}{2}\cos^{-1}\left ( \dfrac{5}{7} \right ) \right ] = \dfrac{14}{5}$$
Question 2
1 / -0

$$\tan ^{ -1 }{ \sqrt { 3 } } -\cot ^{ -1 }{ \left( -\sqrt { 3 } \right) } $$ is equal to

A
$$\pi$$

B
$$-\cfrac{\pi}{2}$$

C
$$0$$

D
$$2\sqrt{3}$$

Solution

$$\tan^-1 \sqrt {3}=\dfrac{\pi}{3}$$
and
$$\cot^-1(-\sqrt{3})=\left(\pi-\dfrac{\pi}{6}\right)=\dfrac{5\pi}{6}$$
$$\therefore$$ $$\dfrac{\pi}{3}-\dfrac{5\pi}{6}=\dfrac{-3\pi}{6}=\dfrac{-\pi}{2}$$
Question 3
1 / -0

$${\cos ^{ - 1}}\left\{ {\dfrac{1}{{\sqrt 2 }}\left( {\cos \dfrac{{9\pi }}{{10}} - \sin \dfrac{{9\pi }}{{10}}} \right)} \right\} = $$

A
$$\frac{{23\pi }}{{20}}$$

B
$$\frac{{7\pi }}{{10}}$$

C
$$\frac{{7\pi }}{{20}}$$

D
$$\frac{{17\pi }}{{20}}$$

Solution

$$\displaystyle \cos^{-1}\left \{ \frac{1}{\sqrt{2}}\left ( \cos\frac{9\pi }{10}-\sin\frac{9\pi }{10} \right ) \right \}$$

$$\displaystyle =\cos^{-1}\left \{ \left ( \frac{1}{\sqrt{2}} \cos \frac{9\pi }{10}-\frac{1}{\sqrt{2}} \sin\frac{9\pi }{10} \right ) \right \}$$

$$\displaystyle = \cos^{-1}\left \{ \cos\left ( \frac{\pi }{4}+\frac{9\pi }{10} \right ) \right \}$$

$$\displaystyle =\cos^{-1}\left \{ \cos\left ( \frac{23\pi }{20} \right ) \right \}$$

$$=\dfrac{23\pi }{20}$$
Question 4
1 / -0

If $${ x }_{ 1 },{ x }_{ 2},{ x }_{ 3}$$ are positive roots of $$ x^{ 3 }-6x^{ 2 }+3px-2p=0\quad (p\in R)$$, then the value of $$\sin ^{ -1 } \left( \frac { 1 }{ { x }_{ 1 } } +\frac { 1 }{ { x }_{ 2 } } \right) +\cos ^{ -1 }{ \left( \frac { 1 }{ { x }_{ 2 } } +\frac { 1 }{ { x }_{ 3 } } \right) } -\tan ^{ -1 }{ \left( \frac { 1 }{ { x }_{ 3 } } +\frac { 1 }{ { x }_{ 1 } } \right) } $$ is equal to

A
$$\frac { \pi }{ 8 } $$

B
$$\frac { \pi }{ 6 } $$

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

D
$$\pi $$

Solution

$$x^3-6x^2+3px-2p=0$$
$$x_1$$
$$x_2$$
$$x_3$$
are the zeros
$$x_1+x_2+x_3=6$$
$$x_1x_2+x_2x_3+x_3x_1=3p$$ $$\left[\because \dfrac{x_1+x_2+x_3}{3}\geq (2p)^{\dfrac{1}{3}}, 2\geq (2p)^{\dfrac{1}{3}}\Rightarrow p\leq 4\right]$$
$$x_1x_2x_3=2p$$
$$\dfrac{3p}{3}\geq ((x_1x_2x_3)^2)^{1/3}$$
$$p^3\geq 4p^2$$
$$p\geq 4$$
$$\Rightarrow p=4$$
$$\therefore x_1x_2+x_2x_3+x_3x_1=12$$
$$x_1x_2x_3=8$$
$$x_1+x_2+x_3=6$$
$$\sin^{-1}(1)+\cos^{-1}(1)-\tan^{-1}(1)$$
$$\dfrac{\pi}{2}+0-\dfrac{\pi}{4}=\dfrac{\pi}{4}$$.
Question 5
1 / -0

If $$\tan{\alpha}=\dfrac{m}{m+1}$$ and $$\tan{\beta}=\dfrac{1}{2m+1}$$ find the possible values of $$(\alpha+\beta)$$

A
$$30^\circ$$

B
$$90^\circ$$

C
$$60^\circ$$

D
$$45^\circ$$

Solution

Given that $$\tan \alpha =\dfrac{m}{m+1}$$ ……. (1) and $$\tan \beta =\dfrac{1}{2m+1}$$ …….. (2)

Find $$\alpha +\beta =?$$

We know that,

$$\tan \left( \alpha +\beta \right)=\dfrac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }$$

Put the value of $$\tan \alpha $$ and $$\tan \beta $$

$$ \tan \left( \alpha +\beta \right)=\dfrac{\dfrac{m}{m+1}+\dfrac{1}{2m+1}}{1-\dfrac{m}{m+1}\dfrac{1}{2m+1}} $$

$$ \Rightarrow \tan \left( \alpha +\beta \right)=\dfrac{\dfrac{m\left( 2m+1 \right)+m+1}{\left( m+1 \right)\left( 2m+1 \right)}}{\dfrac{\left( m+1 \right)\left( 2m+1 \right)-m}{\left( m+1 \right)\left( 2m+1 \right)}} $$

$$ \Rightarrow \tan \left( \alpha +\beta \right)=\dfrac{2{{m}^{2}}+m+m+1}{2{{m}^{2}}+2m+m+1-m} $$

$$ \Rightarrow \tan \left( \alpha +\beta \right)=\dfrac{2{{m}^{2}}+m+m+1}{2{{m}^{2}}+m+m+1} $$

$$ \Rightarrow \tan \left( \alpha +\beta \right)=1 $$

$$ \Rightarrow \tan \left( \alpha +\beta \right)=\tan {{45}^{0}} $$

$$ \Rightarrow \alpha +\beta ={{45}^{0}} $$

Hence, this is the answer
Question 6
1 / -0

$$\mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^n {{{\tan }^{ - 1}}\left( {\dfrac{{2r}}{{1 - {r^2} + {r^4}}}} \right)} $$ is equal to

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

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

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

D
none of these

Solution

$$\lim {n\to \infty } \sum _{ r=1 }^{ n }{ \tan ^{ -1 }{ \left(\dfrac { 2r }{ 1-{ r }^{ 2 }+{ r }^{ 4 } } \right)} } $$

$$=\lim {n\to \infty} \sum _{ r=1 }^{ n }{ \tan ^{ -1 }{ \left(\dfrac { 2r }{ 1+r^{ 2 }({ r }^{ 2 }-1) } \right) } } $$

$$=\lim {n\to \infty } \sum _{ r=1 }^{ n }{ \tan ^{ -1 }{ \left(\dfrac { r(r+1)-r(r-1) }{ 1+[r({ r }-1)][r(r+1)] } \right) } } $$

$$=\lim {n\to \infty } \sum _{ r=1 }^{ n }{ \tan ^{ -1 }{ (r(r+1)- } } \tan ^{ -1 }{ (r(r-1)) } $$

$$=\lim {n\to \infty } [tan ^{-1} (1(1+1)- \tan ^{-1} (1(1-1))+tan^{-1} (2(2+1))-\tan ^{-1} (2(2-1))\\ \quad +.......+\tan^{-1}(n(n+1))-\tan^{-1} (n(n-1))]$$

$$=\lim {n\to \infty } [tan ^{-1} (2)- \tan ^{-1} (0)+tan^{-1} (6))-\tan ^{-1} (2)+.........+\tan^{-1}(n(n+1))-\tan^{-1} (n(n-1))]$$

$$=\lim {n\to \infty } [- \tan ^{-1} (0)+\tan^{-1}(n(n+1))]$$

$$=\lim {n\to \infty } [\tan^{-1}(n(n+1))]$$

$$=\lim {n\to \infty } [\tan^{-1}(n^2+n))]$$

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

$$=\frac {\pi}{2}$$
Question 7
1 / -0

Given that $$0 \le x\le \dfrac 12$$ the value of $$\tan \left[ { sin }^{ -1 }\left( \dfrac { x }{ \sqrt { 2 } } +\sqrt { \dfrac { 1-{ x }^{ 2 } }{ 2 } } \right) -{ sin }^{ -1 }x \right]$$ is

A
$$-1$$

B
$$1$$

C
$$1\sqrt{3}$$

D
$$\sqrt{3}$$

Solution

$$\tan \left[\sin^{-1}\left(\dfrac{x}{\sqrt2} + \sqrt{\dfrac{1-x^2}{2}}\right)-\sin x\right]$$

Put $$x = \sin \theta$$

$$\tan \left[\sin^{-1}\left(\dfrac{\sin \theta + \cos \theta}{\sqrt{2}}\right)+(-\theta)\right]$$

$$= \tan \left[\sin^{-1}\left(\sin \theta \cos\left(\dfrac{\pi}{4} \right)+\cos \theta\sin\left(\dfrac{\pi}{4} \right)\right)-\theta\right]$$

$$= \tan \left[\sin^{-1}\sin\left(\dfrac{\pi}{4} + \theta\right)-\theta\right]$$

$$= \tan \left(\dfrac{\pi}{4}\right)$$

$$= 1$$
Question 8
1 / -0

The value of $$\sin^{-1}\left\{ \cot { \left\{ \sin ^{ -1 }{ \sqrt { \frac { 2-\sqrt { 3 } }{ 4 } } +\cos ^{ }{ \frac { \sqrt { 12 } }{ 4 } +\sec ^{ -1 }{ \sqrt { 2 } } } } \right\} } \right\} $$ is equal to

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

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

C
$$0$$

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

Solution

Solution:-
$$ sin^{-1}\left \{ cot \left \{ sin^{-1}\sqrt{\dfrac{2-\sqrt{3}}{4}}+cos^{-1}\dfrac{\sqrt{12}}{4}+sec^{-1}\sqrt{2} \right \} \right \}$$
We know that
$$ sin \dfrac{\pi}{12} = \sqrt{\dfrac{2-\sqrt{3}}{4}} cos\dfrac{\pi}{6} = \dfrac{\sqrt{3}}{2} = \dfrac{\sqrt{12}}{4}$$
$$ sec \dfrac{\pi}{4} = \sqrt{2}$$
$$ sin^{-1} (cot(\dfrac{\pi}{12}+\dfrac{\pi}{6} + \dfrac{\pi}{4}))$$
$$ sin^{-1} \left \{ cot \left ( \dfrac{\pi}{2} \right ) \right \}$$
$$ sin^{-1}0$$
$$0$$
C is correct.
Question 9
1 / -0

the value of $$\cos ^{ -2 }{ \left( \cos { 10 } \right) }$$ is

A
$$4\pi +10$$

B
$$4\pi -10$$

C
$$2\pi +10$$

D
$$2\pi -10$$

Solution
Question 10
1 / -0

The solution set of the equation $$\sin^{-1}\sqrt{1-x^2}+\cos^{-1}x=\cot^{-1}\dfrac{\sqrt{1-x^2}}{x}-\sin^{-1}x$$ is?

A
$$[-1, 1]-\{0\}$$

B
$$(0, 1] \cup \{-1\}$$

C
$$[-1, 0)\cup \{1\}$$

D
$$[-1, 1]$$

Solution

$$\left( {{{\sin }^{ - 1}}x + {{\cos }^{ - 1}}x} \right) = {\cot ^{ - 1}}\frac{{\sqrt {1 - {x^2}} }}{x} - {\sin ^{ - 1}}\sqrt {1 - {x^2}} $$
let $$x = \cos \theta ,\,\theta = {\cos ^{ - 1}}x$$
\begin{array}{l} \frac { \pi }{ 2 } ={ \cot ^{ -1 } }\frac { { \sqrt { 1-{ { \cos }^{ 2 } }\theta } } }{ { \cos \theta } } -{ \sin ^{ -1 } }\sqrt { 1-{ { \cos }^{ 2 } }\theta } \\ \frac { \pi }{ 2 } ={ \cot ^{ -1 } }\left( { \frac { { \sin \theta } }{ { \cos \theta } } } \right) -{ \sin ^{ -1 } }\left( { \sin \theta } \right) \\ \frac { \pi }{ 2 } ={ \cot ^{ -1 } }\left[ { \cot \left( { \frac { \pi }{ 2 } -\theta } \right) } \right] -\theta \\ \frac { \pi }{ 2 } =\frac { \pi }{ 2 } -\theta -\theta \\ 0=-2\theta \\ \theta =0 \\ { \cos ^{ -1 } }x=0 \\ x=\cos { 0^{ 0 } } \\ x=1 \\ \left[ { -1,1 } \right] -\{ 0\} \end{array}

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Re-Attempt Weekly Quiz Competition

Inverse Trigonometric Functions Test - 35

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Inverse Trigonometric Functions Test - 35

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SHARING IS CARING

Question 1

$$\dfrac{5}{7}$$

$$\dfrac{10}{7}$$

$$\dfrac{14}{5}$$

$$\dfrac{7}{5}$$

Solution

Question 2

$$\pi$$

$$-\cfrac{\pi}{2}$$

$$0$$

$$2\sqrt{3}$$

Solution

Question 3

$$\frac{{23\pi }}{{20}}$$

$$\frac{{7\pi }}{{10}}$$

$$\frac{{7\pi }}{{20}}$$

$$\frac{{17\pi }}{{20}}$$

Solution

Question 4

$$\frac { \pi }{ 8 } $$

$$\frac { \pi }{ 6 } $$

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

$$\pi $$

Solution

Question 5

$$30^\circ$$

$$90^\circ$$

$$60^\circ$$

$$45^\circ$$

Solution

Question 6

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

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

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

none of these

Solution

Question 7

$$-1$$

$$1$$

$$1\sqrt{3}$$

$$\sqrt{3}$$

Solution

Question 8

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

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

$$0$$

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

Solution

Question 9

$$4\pi +10$$

$$4\pi -10$$

$$2\pi +10$$

$$2\pi -10$$

Solution

Question 10

$$[-1, 1]-\{0\}$$

$$(0, 1] \cup \{-1\}$$

$$[-1, 0)\cup \{1\}$$

$$[-1, 1]$$

Solution

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