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Short Quiz - Chapter 2 Class 12 Inverse Trigonometric Functions

Chapter 2 Class 12 Inverse Trigonometric Functions | 5 questions

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Question 1 of 5
Question 1 of 5
NCERT Exemplar
If \(\sin ^{-1} x+\sin ^{-1} y=\frac{\pi}{2}\), then value of \(\cos ^{-1} x+\cos ^{-1} y\) is

Correct option: A

Answer: Solution (A) is the correct answer. Given that \(\sin ^{-1} x+\sin ^{-1} y=\frac{\pi}{2}\).
Therefore, \(\quad\left(\frac{\pi}{2}-\cos ^{-1} x\right)+\left(\frac{\pi}{2}-\cos ^{-1} y\right)=\frac{\pi}{2}\)

$$ \Rightarrow \quad \cos ^{-1} x+\cos ^{-1} y=\frac{\pi}{2} . $$

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Question 2 of 5
NCERT Exemplar
The value of \(\cot \left[\cos ^{-1}\left(\frac{7}{25}\right)\right]\) is

Correct option: D

Answer: Let

$$ \theta=\cos ^{-1}\left(\frac{7}{25}\right) $$


So,

$$ \cos \theta=\frac{7}{25} $$


This means:

$$ \text { base }=7, \quad \text { hypotenuse }=25 $$


Now,

$$ \begin{gathered} \text { perpendicular }=\sqrt{25^2-7^2} \\ =\sqrt{625-49} \\ =\sqrt{576}=24 \end{gathered} $$


Therefore,

$$ \begin{gathered} \cot \theta=\frac{\text { base }}{\text { perpendicular }} \\ \cot \theta=\frac{7}{24} \end{gathered} $$


Answer: D) \(\frac{7}{24}\)

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Question 3 of 5
NCERT Exemplar
The value of \(\cot \left(\sin ^{-1} x\right)\) is

Correct option: D

Answer: Solution (D) is the correct answer. Let \(\sin ^{-1} x=\theta\), then \(\sin \theta=x\)

$$ \begin{aligned} & \Rightarrow \quad \operatorname{cosec} \theta=\frac{1}{x} \quad \Rightarrow \quad \operatorname{cosec}^2 \theta=\frac{1}{x^2} \\ & \Rightarrow \quad 1+\cot ^2 \theta=\frac{1}{x^2} \Rightarrow \cot \theta=\frac{\sqrt{1-x^2}}{x} . \end{aligned} $$

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Question 4 of 5
NCERT Exemplar
The value of \(\sin \left(2 \sin ^{-1}(\cdot 6)\right)\) is

Correct option: B

Answer: Solution (B) is the correct answer. Let \(\sin ^{-1}(\cdot 6)=\theta\), i.e., \(\sin \theta=\cdot 6\).
Now \(\sin (2 \theta)=2 \sin \theta \cos \theta=2(\cdot 6)(\cdot 8)=.96\).

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Question 5 of 5
NCERT Exemplar
The principal value of the expression \(\cos ^{-1}\left[\cos \left(-680^{\circ}\right)\right]\) is

Correct option: A

Answer: \(\cos ^{-1}\left(\cos \left(680^{\circ}\right)\right)=\cos ^{-1}\left[\cos \left(720^{\circ}-40^{\circ}\right)\right]\)

$$ =\cos ^{-1}\left[\cos \left(-40^{\circ}\right)\right]=\cos ^{-1}\left[\cos \left(40^{\circ}\right)\right]=40^{\circ}=\frac{2 \pi}{9} . $$

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