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Question 51
The value of $$\sin^2 48^\circ + \sin^2 42^\circ - \sec^2 30^\circ + \tan^2 60^\circ $$ is equal to:
Solution
$$\sin 42^\circ$$ =Â $$\sin (90 - 42)^\circ$$ =Â $$\cos 48^\circ$$
$$\sin^2 48^\circ + \cos^2 48^\circ - \sec^2 30^\circ + \tan^2 60^\circ$$ = 1 -Â $$\frac{4}{3}$$ + 3 =$$\frac{8}{3}$$
(Since we know that $$sin ^2 \theta + cos ^2\theta = 1, tan 60^o = \sqrt{3}, sec 60^o = \frac{2}{\sqrt{3}}$$)
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