\left[\frac{5}{8}{sec{\ }}^2{60}^\circ-{tan{\ }}^2{60}^\circ+{cos{\ }}^2{45}^\circ\right] is equal to

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Solution

Q.
$[\frac{5}{8} {s e c}^{2} 60 ° - {t a n}^{2} 60 ° + {c o s}^{2} 45 °]$ is equal to

1 $- \frac{5}{3}$

2 $- \frac{1}{2}$

3 0

4 $\frac{1}{4}$

Ans.
(3)

We have,

$[\frac{5}{8} {s e c}^{2} 60 ° - {t a n}^{2} 60 ° + {c o s}^{2} 45 °] = \frac{5}{8} \times (2)^{2} - (\sqrt{3})^{2} + {(\frac{1}{\sqrt{2}})}^{2} = \frac{5}{8} \times 4 - 3 + \frac{1}{2} = \frac{5}{2} - 3 + \frac{1}{2} = \frac{5 - 6 + 1}{2} = 0$

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