For a simple pendulum, equation of tangential motion is

$F_T=-m g \sin \theta$

$\Rightarrow m l \alpha=-m g \sin \theta$

$\Rightarrow \frac{d^2 \theta}{d t^2}+g \frac{l}{l} \sin \theta=0$

Now, $\quad \sin \theta=\theta-\frac{\theta^3}{3 !}+\ldots$

Approximately, $\sin \theta \approx \theta\left(1-\frac{\theta^2}{6}\right)$

Now, replacing $\theta_{ av }^2=\frac{1}{2} \theta_0^2$

We have, $\frac{d^2 \theta}{d t^2}+\frac{g}{l}\left(1-\frac{\theta_0^2}{12}\right) \theta=0$

$\therefore \quad \omega=\frac{g}{l}\left(1-\frac{\theta_0^2}{12}\right)$

So, time period of oscillation is

$T=\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{g}{l}}\left[\left(1-\frac{\theta_0^2}{12}\right)^{-\frac{1}{2}}\right]$

Now, taking binomial approximately

$T \approx 2 \pi \sqrt{\frac{g}{l}} \cdot\left(1+\frac{\theta_l^2}{24}\right)$

Clearly, this is greater than $T_0=2 \pi \sqrt{\frac{g}{l}}$.