1. Acceleration at the extreme position

At the extreme position, the bob momentarily comes to rest, so:

• speed $v=0$
• hence radial (centripetal) acceleration $= \dfrac{v^2}{l}=0$

Only tangential acceleration acts due to gravity: $a_{\text{ext}}=g\sin\theta$

2. Acceleration at the lowest position

At the lowest position, the tangential component of gravity is zero, so tangential acceleration is zero.

Only radial acceleration acts: $a_{\text{low}}=\frac{v^2}{l}$

We need the speed $v$ at the lowest point.

Using conservation of mechanical energy between extreme and lowest positions:

Loss in potential energy $=$ gain in kinetic energy $mg l(1-\cos\theta)=\frac12 mv^2$

So, $v^2=2gl(1-\cos\theta)$

Hence, $a_{\text{low}}=\frac{v^2}{l}=2g(1-\cos\theta)$

3. Given condition: magnitudes of accelerations are equal

$a_{\text{ext}}=a_{\text{low}}$

Therefore, $g\sin\theta=2g(1-\cos\theta)$

Cancel $g$: $\sin\theta=2(1-\cos\theta)$

4. Solve the trigonometric equation

Use half-angle identities: $\sin\theta=2\sin\frac\theta2\cos\frac\theta2$ $1-\cos\theta=2\sin^2\frac\theta2$

Substitute: $2\sin\frac\theta2\cos\frac\theta2=2\cdot 2\sin^2\frac\theta2$ $2\sin\frac\theta2\cos\frac\theta2=4\sin^2\frac\theta2$

For non-zero swing, $\sin\dfrac\theta2\neq 0$, so divide by $2\sin\dfrac\theta2$: $\cos\frac\theta2=2\sin\frac\theta2$

Thus, $\tan\frac\theta2=\frac12$

So, $\frac\theta2=\tan^{-1}\left(\frac12\right)$

Hence, $\theta=2\tan^{-1}\left(\frac12\right)$

5. Option check

This matches:

B: $2\tan^{-1}\left(\frac12\right)$

Final Answer: $\boxed{\text{B}}$