Question

If A = 2 tan^–1(2$\left( \frac { 1 } { 3 } \right)$+ sin^–1$\left( \frac { 3 } { 5 } \right)$, then-

• A. A = B
• B. A < B
• C. A > B
• D. None of these

Answer 1

Answer: (C)

A > B

Answer 2

We have A = 2 tan^–1(2$\sqrt { 2 }$– 1) = 2 tan^–1 (1.828)

Ž A > 2 tan^–1 $\sqrt { 3 }$ Ž A > $\frac { 2 \pi } { 3 }$

Next sin^{– 1}$\left( \frac { 1 } { 3 } \right)$< sin^–1$\left( \frac { 1 } { 3 } \right)$ < $\frac { \pi } { 6 }$

Ž sin^{– 1}$\frac { 1 } { 3 }$< $\frac { \pi } { 2 }$

Also 3 sin^{– 1} $\left( \frac { 1 } { 3 } \right)$= sin^–1 $\left[ 3 \cdot \frac { 1 } { 3 } - 4 \left( \frac { 1 } { 3 } \right) ^ { 3 } \right]$

= sin^–1 $\left( \frac { 23 } { 27 } \right)$ = sin^–1 (0. 852)

Ž 3 sin^–1 $\left( \frac { 1 } { 3 } \right)$ < sin^–1 $\left( \frac { \sqrt { 3 } } { 2 } \right)$ Ž 3 sin^–1$\left( \frac { 1 } { 3 } \right)$ < $\frac { \pi } { 3 }$

Further sin^–1 $\left( \frac { 3 } { 5 } \right)$ = sin^–1 (0. 6) < sin^–1 $\left( \frac { \sqrt { 3 } } { 2 } \right)$

Ž sin^–1 $\left( \frac { 3 } { 5 } \right)$ < $\frac { \pi } { 3 }$

Hence, B = 3 sin^–1$\left( \frac { 1 } { 3 } \right)$+ sin^–1 $\left( \frac { 3 } { 5 } \right)$

< $\frac { \pi } { 3 }$ + $\frac { \pi } { 3 }$ = $\frac { 2 \pi } { 3 }$ . Hence A > B