Inverse Trigonometric Functions Test

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Inverse Trigonometric Functions Test - 39

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Weekly Quiz Competition

Question 1
1 / -0

The value of $$\sin \left(\dfrac {1}{4}\sin^{-1}\dfrac {\sqrt {63}}{8}\right)$$ is

A
$$\dfrac {1}{2}$$

B
$$\dfrac {1}{3}$$

C
$$\dfrac {1}{2\sqrt {2}}$$

D
$$\dfrac {1}{5}$$

Solution

$$\sin \left(\dfrac 14 \sin ^{-1} \dfrac {\sqrt {63}}8\right)$$

Let $$x=\sin ^{-1} \dfrac {\sqrt {63}}8$$

$$\implies \sin x=\dfrac {\sqrt {63}}8$$

$$\implies \cos x=\dfrac {\sqrt {64-63}}{8}=\dfrac 18$$

$$\cos \dfrac x2 =\sqrt {\dfrac {1+\cos x}{2}}$$

$$=\sqrt {\dfrac 12 \left(\dfrac 98\right)}$$

$$\cos \dfrac x2=\dfrac 34$$

$$\implies \cos \dfrac x4 =\sqrt {\dfrac {1+\cos \dfrac x2}2} $$

$$=\sqrt {\dfrac 12 \left(1+\dfrac 34\right)}$$

$$=\sqrt{\dfrac 78}$$

$$\implies \sin \dfrac x4$$

$$=\sqrt{1-\cos ^2\dfrac x4}$$

$$=\sqrt {1-\dfrac 78}$$

$$=\dfrac 1{2\sqrt 2}$$
Question 2
1 / -0

If $$\sin ^{-1}x + \sin ^{-1}y + \sin ^{-1}z = \dfrac{3\pi}{2}$$, then $$(x^{500} + y^{500} + z^{500}) - (x^{501} + y^{501} + z^{501})$$ is

A
$$0$$

B
$$1$$

C
$$2$$

D
$$4$$

Solution

We know that the range of $${\sin}^{-1}{x}$$ is $$\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$$

Given:$${\sin}^{-1}{x}+{\sin}^{-1}{y}+{\sin}^{-1}{z}=\dfrac{3\pi}{2}$$

$$\Rightarrow\,{\sin}^{-1}{x}+{\sin}^{-1}{y}+{\sin}^{-1}{z}=\dfrac{\pi}{2}+\dfrac{\pi}{2}+\dfrac{\pi}{2}$$

$$\Rightarrow\,{\sin}^{-1}{x}=\dfrac{\pi}{2}\,\,\,{\sin}^{-1}{y}=\dfrac{\pi}{2}\,\,\,{\sin}^{-1}{z}=\dfrac{\pi}{2}$$

$$\Rightarrow\,x=\sin{\dfrac{\pi}{2}},\,\,y=\sin{\dfrac{\pi}{2}},\,\,z=\sin{\dfrac{\pi}{2}}$$

$$\Rightarrow\,x=1,y=1,z=1$$

$$\left({x}^{500}+{y}^{500}+{z}^{500}\right)-\left({x}^{501}+{y}^{501}+{z}^{501}\right)$$

$$=\left({1}^{500}+{1}^{500}+{1}^{500}\right)-\left({1}^{501}+{1}^{501}+{1}^{501}\right)$$

$$=\left(1+1+1\right)-\left(1+1+1\right)=3-3=0$$
Question 3
1 / -0

If $$\tan^{-1}\left(\dfrac{x+1}{x-1}\right)+\tan^{-1}\left(\dfrac{x-1}{x}\right)=\pi +\tan^{-1}(-7)$$, then $$x=$$

A
$$2$$

B
$$-2$$

C
$$1$$

D
No solution

Solution
Question 4
1 / -0

The number of real solutions of $$\cos ^ { - 1 } x + \cos ^ { - 1 } 2 x = - \pi$$ is

A
$$0$$

B
$$1$$

C
$$2$$

D
None

Solution

$$\cos ^ { - 1 } x + \cos ^ { - 1 } 2 x = - \pi$$
We know that, range of $$\cos^{-1}x$$ is $$[0,\pi]$$ and range of $$\cos^{-1}2x$$ is $$[0,\pi]$$.
So, here we can see both ranges are in positive quantity.
So, sum of positive quantity cannot be negative quantity.
$$\therefore$$ $$\cos ^ { - 1 } x + \cos ^ { - 1 } 2 x = - \pi$$ is not possible.
$$\therefore$$ The number of real solution is $$0.$$
Question 5
1 / -0

The value of $$ \sin (\cot^{-1}x)$$ is

A
$$\sqrt{1+x^2}$$

B
$$x$$

C
$$(1+x^2)^{-3/2}$$

D
$$(1+x^2)^{-1/2}$$

Solution
Question 6
1 / -0

$$\cos ^ { - 1 } \left( \cos \left( \dfrac { - 17 \pi } { 5 } \right) \right)$$ is equal to

A
$$- \dfrac { 17 \pi } { 5 }$$

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

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

D
none of these

Solution

$$\cos^{-1}\left(\cos \left(\dfrac{-17\pi}{5}\right)\right)$$

$$\Rightarrow \cos^{-1}\left(\cos\dfrac{17\pi}{5}\right)$$

$$\Rightarrow \cos^{-1}\left(\cos\left(2\pi +\dfrac{7\pi}{5}\right)\right)$$

$$\Rightarrow \cos^{-1}\left(\cos\dfrac{7\pi}{5}\right)$$

$$\Rightarrow \cos^{-1}\left(\cos\left(2\pi -\dfrac{3\pi}{5}\right)\right)$$

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

$$=\dfrac{3\pi}{5}$$.
Question 7
1 / -0

The value of $$\tan ^{-1} \left (\dfrac{1}{3}\right) +\tan ^{-1} \left (\dfrac{2}{9}\right) + \tan ^{-1}\left (\dfrac{4}{33}\right) + \tan ^{-1} \left (\dfrac{8}{129}\right) +$$.......$$n$$ terms is

A
$$\tan^{-1} 2^n - \dfrac{\pi}{4}$$

B
$$\tan^{-1} 2^n$$

C
$$\cot^{-1} 2^n$$

D
$$\dfrac{\sin^{-1} 2^n}{\cos^{-1} 2^n}$$

Solution

$$\tan^{-1} \left(\dfrac{1}{3} \right) + \tan^{-1} \left(\dfrac{2}{9} \right) + \tan^{-1} \left(\dfrac{4}{33} \right) + \tan^{-1} \left(\dfrac{8}{129} \right) + .... n$$ terms.

$$= \tan^{-1} \left(\dfrac{2 - 1}{1 + 2.1} \right) + \tan^{-1} \left(\dfrac{4 - 2}{1 + 4.2} \right) + \tan^{-1} \left(\dfrac{8 - 4}{1 + 8.4} \right) + \tan^{-1} \left(\dfrac{16 - 8}{1 + 16 \times 8} \right) +...+ \tan^{-1} \left(\dfrac{2^n - 2^{n - 1}}{1 + 2^n \times 2^{n - 1}} \right)$$

we know that $$\tan^{-1}x-\tan^{-1}y=\tan^{-1}{\dfrac{x-y}{1+xy}}$$

$$= \tan^{-1} 2 - \tan^{-1} 1 + \tan^{-1} 4 - \tan^{-1} 2 + \tan^{-1} 8 - \tan^{-1} 4 + \tan^{-1} 16 - \tan^{-1} 8 + ... + \tan^{-1} 2^n - \tan^{-1} 2^{n - 1}$$

$$= \tan^{-1} 2^n - \tan^{-1} 1$$

$$= \tan^{-1} 2^n - \dfrac{\pi}{4}$$
Question 8
1 / -0

$$\sinh { \left( \cosh ^{ -1 }{ x } \right) } =$$

A
$$\sqrt{x^{2}+1}$$

B
$$\dfrac{1}{\sqrt{x^{2}+1}}$$

C
$$\sqrt{x^{2}-1}$$

D
$$\dfrac{1}{\sqrt{x^{2}-1}}$$

Solution

We have,
$$\sin h (\cos h^{-1}x)=?$$
Let
$$\cos h^{-1}x=A--------------(1)$$
$$\cos h\ A=x$$
Squaring both side
$$\cos h^{2} A=x^{2}$$
$$1+\sin h^{2} A=x^{2}$$
$$\because \cos h^{2}x- \sin h^{2} x=1$$
$$\sin h^{2} A=x^{2}-1$$
$$\sin h\ A= \sqrt{x^{2}-1}$$
$$A= \sin h^{-1} \sqrt{x^{2}-1} ---------(2)$$
By equation $$(1)$$ and $$(2)$$ to
$$\cos h^{-1} x= \sin h^{-1} \sqrt{x^{2}-1}$$
According to given function.
$$\sin h (\cos h^{-1} x)$$
$$\Rightarrow \sin h \sin h^{-1} \sqrt{x^{2}-1}$$
$$\Rightarrow \sqrt{x^{2}-1}$$
Hence, this is the answer.
Question 9
1 / -0

$$\sin\ h^{-1}{\left(2^{3/2}\right)}$$=

A
$$\log \left(2+\sqrt{8}\right)$$

B
$$\log \left(3+\sqrt{8}\right)$$

C
$$\log \left(2-\sqrt{8}\right)$$

D
$$\log \left(\sqrt{8}+\sqrt{7}\right)$$

Solution

$$\sin h^{-1}(2^{3/2})=?$$
Now, we know that
$$\sin h^{-1}x=\log{(x+\sqrt{x^2+1})}$$
then,
$$\sin h^{-1}(2^{3/2})=\log \left(2^{3/2}+\sqrt{(2^{3/2})^{2}+1}\right)$$
$$=\log{\left(\sqrt{2^3}+\sqrt{2^3+1}\right)}$$
$$=\log{(\sqrt{8}+\sqrt{9})}$$
$$=\log{(\sqrt{9}+\sqrt{8})}$$
$$=\log{(\sqrt{3^2}+\sqrt{8})}$$
$$\sin h^{-1}(2^{3/2})=\log{(3+\sqrt{8})}$$
Hence, this is the answer.
Question 10
1 / -0

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

A
$$\tan^2 \left (\dfrac{\alpha}{2}\right)$$

B
$$\cot^2 \left (\dfrac{\alpha}{2}\right)$$

C
$$\tan \alpha$$

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

Solution

Given equation: $${ cot }^{ -1 }(\sqrt { cos\alpha } )-{ tan }^{ -1 }(\sqrt { cos\alpha } )=x$$

$$let\quad { cot }^{ -1 }(\sqrt { cos\alpha } )=P$$

$$ \Longrightarrow cotP=\sqrt { cos\alpha } $$

$$ \Longrightarrow sinP=\dfrac { 1 }{ \sqrt { 1+cos\alpha } } $$

Also $${ tan }^{ -1 }(\sqrt { cos\alpha } )=Q$$

$$ \Longrightarrow tanQ=\sqrt { cos\alpha } $$

$$ \Longrightarrow sinQ=\dfrac { \sqrt { cos\alpha } }{ \sqrt { 1+cos\alpha } } $$

Now the given equation can be written as

$$ x=P-Q$$

$$ sinx=sin(P-Q)$$

$$ sinx=sinPcosQ-cosPsinQ$$

$$ sinx=\left( \dfrac { 1 }{ \sqrt { 1+cos\alpha } } \right) \left( \dfrac { 1 }{ \sqrt { 1+cos\alpha } } \right) -\left( \dfrac { \sqrt { cos\alpha } }{ \sqrt { 1+cos\alpha } } \right) \left( \dfrac { \sqrt { cos\alpha } }{ \sqrt { 1+cos\alpha } } \right) $$

$$sinx=\dfrac { 1-cos\alpha }{ 1+cos\alpha } =\dfrac { 2{ sin }^{ 2 }\dfrac { \alpha }{ 2 } }{ 2{ cos }^{ 2 }\dfrac { \alpha }{ 2 } } \\ sinx={ tan }^{ 2 }\dfrac { \alpha }{ 2 }$$

Hence option (a) is correct.

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Inverse Trigonometric Functions Test - 39

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Inverse Trigonometric Functions Test - 39

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SHARING IS CARING

Question 1

$$\dfrac {1}{2}$$

$$\dfrac {1}{3}$$

$$\dfrac {1}{2\sqrt {2}}$$

$$\dfrac {1}{5}$$

Solution

Question 2

$$0$$

$$1$$

$$2$$

$$4$$

Solution

Question 3

$$2$$

$$-2$$

$$1$$

No solution

Solution

Question 4

$$0$$

$$1$$

$$2$$

None

Solution

Question 5

$$\sqrt{1+x^2}$$

$$x$$

$$(1+x^2)^{-3/2}$$

$$(1+x^2)^{-1/2}$$

Solution

Question 6

$$- \dfrac { 17 \pi } { 5 }$$

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

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

none of these

Solution

Question 7

$$\tan^{-1} 2^n - \dfrac{\pi}{4}$$

$$\tan^{-1} 2^n$$

$$\cot^{-1} 2^n$$

$$\dfrac{\sin^{-1} 2^n}{\cos^{-1} 2^n}$$

Solution

Question 8

$$\sqrt{x^{2}+1}$$

$$\dfrac{1}{\sqrt{x^{2}+1}}$$

$$\sqrt{x^{2}-1}$$

$$\dfrac{1}{\sqrt{x^{2}-1}}$$

Solution

Question 9

$$\log \left(2+\sqrt{8}\right)$$

$$\log \left(3+\sqrt{8}\right)$$

$$\log \left(2-\sqrt{8}\right)$$

$$\log \left(\sqrt{8}+\sqrt{7}\right)$$

Solution

Question 10

$$\tan^2 \left (\dfrac{\alpha}{2}\right)$$

$$\cot^2 \left (\dfrac{\alpha}{2}\right)$$

$$\tan \alpha$$

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

Solution

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