SATHEE: Inverse Circular Functions 2 Question 4

Inverse Circular Functions 2 Question 4

4. If $α = 3 \sin^{- 1} \frac{6}{11}$ and $β = 3 \cos^{- 1} \frac{4}{9}$ , where the inverse trigonometric functions take only the principal values, then the correct option(s) is/are

(2015 Adv.)

(a) $\cos β > 0$

(b) $\sin β < 0$

(d) $\cos α < 0$

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Answer:

Correct Answer: 4. (c)

Solution:

Here, $α = 3 \sin^{- 1} \frac{6}{11}$ and $β = 3 \cos^{- 1} \frac{4}{9}$ as $\frac{6}{11} > \frac{1}{2}$

$\begin{aligned} \Rightarrow \sin^{- 1} \frac{6}{11} > \sin^{- 1} \frac{1}{2} = \frac{π}{6} \\ ∴ α = 3 \sin^{- 1} \frac{6}{11} > \frac{π}{2} \\ \Rightarrow \cos α < 0 \end{aligned}$

Now, $β = 3 \cos^{- 1} \frac{4}{9}$

As $\frac{4}{9} < \frac{1}{2} \Rightarrow \cos^{- 1} \frac{4}{9} > \cos^{- 1} \frac{1}{2} = \frac{π}{3}$

$∴ β = 3 \cos^{- 1} \frac{4}{9} > π$

$∴ \cos β < 0$ and $\sin β < 0$

Now, $α + β$ is slightly greater than $\frac{3 π}{2}$ .

$∴ \cos (α + β) > 0$

Inverse Circular Functions 2 Question 4

4. If $α = 3 \sin^{- 1} \frac{6}{11}$ and $β = 3 \cos^{- 1} \frac{4}{9}$ , where the inverse trigonometric functions take only the principal values, then the correct option(s) is/are

Answer:

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Inverse Circular Functions 2 Question 4

4. If α=3sin−1⁡611 and β=3cos−1⁡49, where the inverse trigonometric functions take only the principal values, then the correct option(s) is/are

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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4. If $α = 3 \sin^{- 1} \frac{6}{11}$ and $β = 3 \cos^{- 1} \frac{4}{9}$ , where the inverse trigonometric functions take only the principal values, then the correct option(s) is/are