Question 55

The value of

$$ \sin^2 42^\circ + \sin^2 48^\circ + \tan^2 60^\circ - \cosec 30^\circ$$ is equal to:

Solution

$$ \sin^2 42^\circ + \sin^2 48^\circ + \tan^2 60^\circ - \cosec 30^\circ$$

= $$ \sin^2 42^\circ + \cos^2 42^\circ + \tan^2 60^\circ - \cosec 30^\circ$$

= 1 + 3 - 2 = 2

{Using $$\sin^2\theta + \cos^2\theta = 1, \tan60\degree = \sqrt{3} , \cosec30\degree = 2$$}

So , the answer would be option d)2.

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