The greater of the two angles A = 2...

MathematicsProperties of Inverse Trigonometric FunctionsJEE Advanced 1989Moderate

The greater of the two angles $A = 2 tan^{- 1} (22 - 1)$ and $B = 3 sin^{- 1} (1/3) + sin^{- 1} (3/5)$ is .........

Answer:A

Visualized Solution (Hindi)

Defining the Challenge

We need to compare two angles:
$A = 2 tan^{- 1} (22 - 1)$
$B = 3 sin^{- 1} (\frac{1}{3}) + sin^{- 1} (\frac{3}{5})$
Strategy: Compare both angles with the reference value $\frac{2 π}{3}$ .

Estimating Angle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

Evaluate the argument of $tan^{- 1}$ in $A$ :
$2 \approx 1.414 \Rightarrow 22 - 1 \approx 2 (1.414) - 1 = 1.828$
Compare with $3$ :
$3 \approx 1.732$
Since $1.828 > 1.732$ , then $tan^{- 1} (1.828) > tan^{- 1} (3) = \frac{π}{3}$
Multiplying by $2$ : $A > 2 \times \frac{π}{3} \Rightarrow A > \frac{2 π}{3}$

The Logic Bridge for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B

To simplify $B$ , use the identity for $3 sin^{- 1} x$ :
$3 sin^{- 1} x = sin^{- 1} (3 x - 4 x^{3})$
Substitute $x = \frac{1}{3}$ into the formula.

Computing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 3 sin^{- 1} (\frac{1}{3})

Calculation for the first term of $B$ :
$3 sin^{- 1} (\frac{1}{3}) = sin^{- 1} [3 (\frac{1}{3}) - 4 (\frac{1}{3})^{3}]$
$3 sin^{- 1} (\frac{1}{3}) = sin^{- 1} [1 - \frac{4}{27}] = sin^{- 1} (\frac{23}{27})$
Numerical approximation: $\frac{23}{27} \approx 0.852$

Estimating Angle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B

Compare terms of $B$ with $\frac{π}{3}$ :
Reference: $sin (\frac{π}{3}) = \frac{3}{2} \approx 0.866$
Term 1: $sin^{- 1} (0.852) < sin^{- 1} (0.866) = \frac{π}{3}$
Term 2: $sin^{- 1} (0.6) < sin^{- 1} (0.866) = \frac{π}{3}$
Therefore: $B < \frac{π}{3} + \frac{π}{3} \Rightarrow B < \frac{2 π}{3}$

Final Comparison

From our analysis:
1. $A > \frac{2 π}{3}$
2. $B < \frac{2 π}{3}$
Conclusion: $A > B$
The greater of the two angles is A.

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