SATHEE: Inverse Circular Functions 1 Question 2

Inverse Circular Functions 1 Question 2

2. The greater of the two angles $A = 2 \tan^{- 1} (2 \sqrt{2} - 1)$ and $B = 3 \sin^{- 1} \frac{1}{3} + \sin^{- 1} \frac{3}{5}$ is …… .

$(1989, 2 M)$

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Answer:

Correct Answer: 2. $(A)$

Solution:

Given, $A = 2 \tan^{- 1} (2 \sqrt{2} - 1)$

and $B = 3 \sin^{- 1} \frac{1}{3} + \sin^{- 1} \frac{3}{5}$

Here, $A = 2 \tan^{- 1} (2 \sqrt{2} - 1)$

$= 2 \tan^{- 1} (2 \times 1.414 - 1)$

$= 2 \tan^{- 1} (1.828)$

$∴ A > 2 \tan^{- 1} (\sqrt{3}) = 2 \cdot \frac{π}{3} = \frac{2 π}{3}$

To find the value of $B$ , we first say

$\sin^{- 1} \frac{1}{3} < \sin^{- 1} \frac{1}{2} = \frac{π}{6}$

so that $0 < 3 \sin^{- 1} \frac{1}{3} < \frac{π}{2}$

Now, $3 \sin^{- 1} \frac{1}{3} = \sin^{- 1} 3 \cdot \frac{1}{3} - 4 \cdot \frac{1}{27}$

$\begin{aligned} = \sin^{- 1} \frac{23}{27} \\ = \sin^{- 1} (0.851) < \sin^{- 1} \frac{\sqrt{3}}{2} = \frac{π}{3} \\ \sin^{- 1} \frac{3}{5} = \sin^{- 1} (0.6) < \sin^{- 1} \frac{\sqrt{3}}{2} = \frac{π}{3} \\ ∴ B < \frac{π}{3} + \frac{π}{3} = \frac{2 π}{3} \end{aligned}$

Thus, $A > \frac{2 π}{3}$ and $B < \frac{2 π}{3}$

Hence, greater angle is $A$ .

Inverse Circular Functions 1 Question 2

2. The greater of the two angles $A = 2 \tan^{- 1} (2 \sqrt{2} - 1)$ and $B = 3 \sin^{- 1} \frac{1}{3} + \sin^{- 1} \frac{3}{5}$ is …… .

Answer:

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Inverse Circular Functions 1 Question 2

2. The greater of the two angles A=2tan−1⁡(22−1) and B=3sin−1⁡13+sin−1⁡35 is …… .

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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2. The greater of the two angles $A = 2 \tan^{- 1} (2 \sqrt{2} - 1)$ and $B = 3 \sin^{- 1} \frac{1}{3} + \sin^{- 1} \frac{3}{5}$ is …… .