Inverse Trigonometric Functions Test

Result

Inverse Trigonometric Functions Test - 68

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Weekly Quiz Competition

Question 1
1 / -0

$$tan^{-1}(1-x^{2}-\frac{1}{x^{2}})+sin^{-1}(x^{2}+\frac{1}{x^{2}}-1)$$ is equal to

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

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

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

D
$$\pi $$

Solution
Question 2
1 / -0

Evaluate : $$\sin \left(\dfrac {1}{2}\cos^{-1}\dfrac {4}{5}\right)$$

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

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

C
$$\dfrac {1}{\sqrt {10}}$$

D
$$\dfrac {2}{\sqrt {10}}$$

Solution
Question 3
1 / -0

If $$cot^{1}\frac{1}{x}+cot^{-1}\frac{1}{y}+cot^{-1 }\frac{1}{z}=\frac{\pi }{2}$$ then.....

A
x+y+z=xyz

B
xy+yz+zx=1

C
x+y+z=1

D
$$x^{2}+y^{2}+z^{2}=1$$

Solution
Question 4
1 / -0

If $$\alpha =2\tan^{-1}(\sqrt{3-2\sqrt{2}})+\sin^{-1}\left(\dfrac{1}{\sqrt{6}-\sqrt{2}}\right), \beta =\cot^{-1}(\sqrt{3}-2)+\dfrac{1}{8}\sec^{-1}(-2)$$ & $$\gamma =\tan^{-1}\dfrac{1}{\sqrt{2}}+\cos^{-1}\dfrac{1}{\sqrt{3}}$$, then?

A
$$\alpha =\beta$$

B
$$\alpha +\beta =3\gamma$$

C
$$4(\beta -\gamma)=\alpha$$

D
$$\beta =\gamma$$
Question 5
1 / -0

If $$\cos^{-1} x > \sin^{-1} x$$, then find the range of x

A
$$-1 \leq x < \dfrac {1}{\sqrt {2}}$$

B
$$0\leq < \dfrac {1}{\sqrt {2}}$$

C
$$\dfrac {1}{\sqrt {2}} < x \leq 1$$

D
$$x > 0$$

Solution

We have $${\cos}^{-1}{x}>{\sin}^{-1}{x}$$

$$\Rightarrow\,\dfrac{\pi}{2}-{\sin}^{-1}{x}>{\sin}^{-1}{x}$$

$$\Rightarrow\,\dfrac{\pi}{2}>2{\sin}^{-1}{x}$$

$$\Rightarrow\,\dfrac{\pi}{4}>{\sin}^{-1}{x}$$

$$\Rightarrow\,{\sin}^{-1}{x}<\dfrac{\pi}{4}$$ .......$$(1)$$

But $$-\dfrac{\pi}{2}\le {\sin}^{-1}{x}<\dfrac{\pi}{2}$$ ........$$(2)$$

From $$(1)$$ and $$(2)$$ we have

$$-\dfrac{\pi}{2}\le {\sin}^{-1}{x}<\dfrac{\pi}{4}$$

$$\Rightarrow\,\sin{\left(-\dfrac{\pi}{2}\right)}\le x<\sin{\dfrac{\pi}{4}}$$

$$\Rightarrow\,-1\le x<\dfrac{1}{\sqrt{2}}$$
Question 6
1 / -0

$$4 sin^{-1}x+cos^{-1} x=\pi $$ then x=

A
$$\frac{-1}{4}$$

B
$$\frac{1}{4}$$

C
$$\frac{-1}{2}$$

D
$$\frac{1}{2}$$

Solution
Question 7
1 / -0

$$ tan^{-1} \left (\dfrac{c_1\,x - y}{c_1\,y + x} \right ) + tan^{-1} \left ( \dfrac{c_2 - c_1}{1 + c_2c_1} \right ) + tan^{-1} \left (\dfrac{c_3 - c_2}{1 + c_3c_2} \right ) + $$ ..... $$ + tan^{-1}\left (\dfrac{1}{c_n} \right ) = $$

A
$$ tan^{-1} \,(y/x) $$

B
$$ tan^{-1}\dfrac{x}{y} $$

C
$$ -tan^{-1}\left ( \dfrac{x}{y} \right ) $$

D
none of these

Solution
Question 8
1 / -0

The value of $$\tan^{-1}\Bigg(\dfrac{x\cos\theta}{1-x\sin\theta}\Bigg)-\cot^{-1}\Bigg(\dfrac{\cos\theta}{x-\sin\theta}\Bigg)$$ is

A
$$2\theta$$

B
$$\theta$$

C
$$\dfrac{\theta}{2}$$

D
independent of $$\theta$$

Solution
Question 9
1 / -0

Let $$\begin{vmatrix}tan^{-1}x & tan^{-1}2x & tan^{-1}3x\\ tan^{-1}3x & tan^{-1}x & tan^{-1}2x\\
tan^{-1}2x & tan^{-1}3x & tan^{-1}x \end{vmatrix}$$= 0, then the number of values of x satisfying the equation is

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution
Question 10
1 / -0

If $$\cot^{-1}\left(\dfrac {-1}{5}\right)=x$$ then $$\sin x=$$ ?

A
$$\dfrac {1}{\sqrt {26}}$$

B
$$\dfrac {5}{\sqrt {26}}$$

C
$$\dfrac {1}{\sqrt {24}}$$

D
$$none\ of\ these$$

Solution