The value of $$\sin^2 20^\circ + \sin^2 70^\circ - \tan^2 45^\circ + \sec 60^\circ$$ is equal to:

Given that,

$$\sin^2 20^\circ + \sin^2 70^\circ - \tan^2 45^\circ + \sec 60^\circ$$

We know that, $$\cos(90-\theta)=\sin \theta$$

$$\sin^2 20^\circ + \cos^2 (90^\circ-70^\circ) - \tan^2 45^\circ + \sec 60^\circ$$

$$\Rightarrow \sin^2 20^\circ + \cos^2 20^\circ - \tan^2 45^\circ + \sec 60^\circ$$

We know that $$\sin^2 \theta +\cos^2 \theta=1$$

So,

$$\Rightarrow 1 - \tan^2 45^\circ + \sec 60^\circ$$

We know that $$\tan 45=1$$ and $$\cos 60^\circ=\dfrac{1}{2}$$

Now, substituting the values

$$\Rightarrow= 1 - 1+ \dfrac{1}{\cos 60^\circ}$$

$$\Rightarrow= 1 - 1+ \dfrac{2}{1}$$

$$\Rightarrow= 2$$