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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
The value of \(\sin \left(2 \tan ^{-1}(.75)\right)\) is equal to

Correct option: C

Answer: $$ \sin \left(2 \tan ^{-1}(0.75)\right) $$


Let

$$ \theta=\tan ^{-1}(0.75)=\tan ^{-1}\left(\frac{3}{4}\right) $$


So,

$$ \tan \theta=\frac{3}{4} $$


Use identity:

$$ \begin{gathered} \sin 2 \theta=\frac{2 \tan \theta}{1+\tan ^2 \theta} \\ =\frac{2 \cdot \frac{3}{4}}{1+\left(\frac{3}{4}\right)^2} \\ =\frac{\frac{3}{2}}{1+\frac{9}{16}} \\ =\frac{\frac{3}{2}}{\frac{25}{16}} \\ =\frac{3}{2} \times \frac{16}{25} \\ =\frac{24}{25}=0.96 \end{gathered} $$


Answer: (C) 0.96

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Question 2 of 5
NCERT Exemplar
The value of \(\tan ^2\left(\sec ^{-1} 2\right)+\cot ^2\left(\operatorname{cosec}^{-1} 3\right)\) is

Correct option: B

Answer: Solution (B) is the correct answer.

$$ \begin{aligned} & \tan ^2\left(\sec ^{-1} 2\right)+\cot ^2\left(\operatorname{cosec}^{-1} 3\right)=\sec ^2\left(\sec ^{-1} 2\right)-1+\operatorname{cosec}^2\left(\operatorname{cosec}^{-1} 3\right)-1 \\ & =2^2 \times 1+3^2-2=11 . \end{aligned} $$

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Question 3 of 5
NCERT Exemplar
If \(\cos \left(\sin ^{-1} \frac{2}{5}+\cos ^{-1} x\right)=0\), then \(x\) is equal to

Correct option: B

Answer: We know:

$$ \cos 90^{\circ}=0 $$


So,

$$ \sin ^{-1} \frac{2}{5}+\cos ^{-1} x=\frac{\pi}{2} $$


Now use identity:

$$ \sin ^{-1} a+\cos ^{-1} a=\frac{\pi}{2} $$


Comparing,

$$ \sin ^{-1} \frac{2}{5}+\cos ^{-1} \frac{2}{5}=\frac{\pi}{2} $$


So,

$$ x=\frac{2}{5} $$


Answer: B) \(\frac{2}{5}\)

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Question 4 of 5
NCERT Exemplar
The domain of the function \(y=\sin ^{-1}\left(-x^2\right)\) is

Correct option: C

Answer: \(y=\sin ^{-1}\left(-x^2\right) \Rightarrow \sin y=-x^2\)
i.e. \(-1 \leq-x^2 \leq 1 \quad(\) since \(-1 \leq \sin y \leq 1)\)
\(\Rightarrow 1 \geq x^2 \geq-1\)

$$ \Rightarrow 0 \leq x^2 \leq 1 $$


$$ \Rightarrow|x| \leq 1 \text { i.e. }-1 \leq x \leq 1 $$

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Question 5 of 5
NCERT Exemplar
If \(\tan ^{-1} x+\tan ^{-1} y=\frac{4 \pi}{5}\), then \(\cot ^{-1} x+\cot ^{-1} y\) equals

Correct option: A

Answer: Given:

$$ \tan ^{-1} x+\tan ^{-1} y=\frac{4 \pi}{5} $$


We need:

$$ \cot ^{-1} x+\cot ^{-1} y $$


For positive values,

$$ \cot ^{-1} x=\frac{\pi}{2}-\tan ^{-1} x $$


So,

$$ \begin{gathered} \cot ^{-1} x+\cot ^{-1} y \\ =\left(\frac{\pi}{2}-\tan ^{-1} x\right)+\left(\frac{\pi}{2}-\tan ^{-1} y\right) \\ =\pi-\left(\tan ^{-1} x+\tan ^{-1} y\right) \\ =\pi-\frac{4 \pi}{5} \\ =\frac{\pi}{5} \end{gathered} $$


Answer: (A) \(\frac{\pi}{5}\)

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