The value of  $$\left[\frac{\sin^2 24^\circ + \sin^2 66^\circ}{\cos^2 24^\circ + \cos^2 66^\circ} + \sin^2 61^\circ + \cos 61^\circ \sin 29^\circ\right]$$ is:

As per the given question,

$$\left[\frac{\sin^2 24^\circ + \sin^2 66^\circ}{\cos^2 24^\circ + \cos^2 66^\circ} + \sin^2 61^\circ + \cos 61^\circ \sin 29^\circ\right]$$

We know that $$\sin(90^\circ-\theta=\cos \theta$$ and $$\cos(90^\circ-\theta)=\sin\theta$$, so this property in the above equation,

$$\Rightarrow \left[\frac{\sin^2 24^\circ + (\cos (90^\circ-66^\circ)^2}{\cos^2 24^\circ + (\sin (90^\circ-66^\circ)^2} + \sin^2 61^\circ + \cos 61^\circ \cos(90^\circ- 29^\circ)\right]$$

$$\Rightarrow \left[\frac{\sin^2 24^\circ + \cos^2 24^\circ}{\cos^2 24^\circ + \sin^2 24^\circ} + \sin^2 61^\circ + \cos 61^\circ \cos61^\circ\right]$$

We know that $$\sin^2\theta+\cos^2 \theta=1$$, so using this property in the above equation,

$$\Rightarrow \left[1 + \sin^2 61^\circ + \cos^2 61^\circ \right]$$

$$\Rightarrow 1 + 1=2$$