The value of
$$ \sin^2 42^\circ + \sin^2 48^\circ + \tan^2 60^\circ - \cosec 30^\circ$$ is equal to:
$$ \sin^2 42^\circ + \sin^2 48^\circ + \tan^2 60^\circ - \cosec 30^\circ$$
=Â $$ \sin^2 42^\circ + \cos^2 42^\circ + \tan^2 60^\circ - \cosec 30^\circ$$
= 1 + 3 - 2 = 2
{Using $$\sin^2\theta + \cos^2\theta = 1, \tan60\degree = \sqrt{3} , \cosec30\degree = 2$$}Â
So , the answer would be option d)2.
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