Inverse Trigonometric Functions Test

Result

Inverse Trigonometric Functions Test - 57

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Weekly Quiz Competition

Question 1
1 / -0

$$Sin h (cos h ^{-1} x) =$$

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
Question 2
1 / -0

The value of $$e^{sinh^{-1} (tan \theta)}$$ is equal to

A
$$cosec \theta + cot \theta$$

B
$$sec \theta + tan \theta$$

C
$$cosec \theta + sec \theta$$

D
$$tan \theta + cot \theta$$

Solution
Question 3
1 / -0

$$\tanh^{1}\left(\dfrac {1}{3}\right)+\coth^{1}(3)=$$.....

A
$$\log 2$$

B
$$\log 3$$

C
$$\log \sqrt {3}$$

D
$$\log \sqrt {2}$$

Solution
Question 4
1 / -0

For which value of x, $$\sin [\cot^{-1}(x+1)]=\cos (\tan^{-1}x)$$.

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

B
$$0$$

C
$$1$$

D
$$\dfrac{-1}{2}$$

Solution
Question 5
1 / -0

If $$cot^{-1} x - cot^{-1} (x+2) = 15^0$$ then x is equal to

A
$$\sqrt{3}$$

B
$$-\sqrt{3}$$

C
$$\sqrt{3} +2$$

D
$$-\sqrt{3} +2$$
Question 6
1 / -0

If $$y=\dfrac{1}{2}\csc\ h^{-1}{\left(\dfrac{1}{2x\sqrt{1+x^{2}}}\right)}$$ then $$x$$=

A
coshy

B
sinhy

C
tanhy

D
cothy

Solution
Question 7
1 / -0

Solve : $$sin h^{-1} (\dfrac{x}{\sqrt{1-x^2}}) =$$

A
$$cos h^{-1} x$$

B
$$-cos h^{-1} x$$

C
$$tan h^{-1} x$$

D
$$-tan h^{-1} x$$

Solution
Question 8
1 / -0

The value of $$\tan^{-1}\left[\dfrac{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}}\right],\left|x\right|<\dfrac{1}{2},x=0$$, is equal to:

A
$$\dfrac{pi}{4}-\cos^{-1}x^{2}$$

B
$$\dfrac{pi}{4}+\dfrac{1}{2}\cos^{-1}x^{2}$$

C
$$\dfrac{pi}{4}+\cos^{-1}x^{2}$$

D
$$\dfrac{pi}{4}-\dfrac{1}{2}\cos^{-1}x^{2}$$

Solution
Question 9
1 / -0

The value of $$\tan{\left\{\dfrac{\pi}{4}+\dfrac{1}{2}\cos^{-1}{(\dfrac{x}{y})}\right\}}+\tan{\left\{\dfrac{\pi}{4}-\dfrac{1}{2}\cos^{-1}{(\dfrac{x}{y})}\right\}}$$

A
$$\dfrac{x}{y}$$

B
$$\dfrac{y}{x}$$

C
$$\dfrac{2y}{x}$$

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

Solution
Question 10
1 / -0

If $$f(x)={ sin }^{ -1 }\left( \frac { \sqrt { 3 } }{ 2 } x-\frac { 1 }{ 2 } \sqrt { 1-{ x }^{ 2 } } \right) -\frac { 1 }{ 2 } \le x\le 1$$, then f(x) is equal to :

A
$${ sin }^{ -1 }\left( \frac { 1 }{ 2 } \right) -{ sin }^{ -1 }(x)$$

B
$${ sin }^{ -1 }x-\frac { \pi }{ 6 } $$

C
$${ sin }^{ -1 }x+\frac { \pi }{ 6 } $$

D
None of these

Solution