2(cos² 45 $°$ + tan² $°$ 60) $-$ 6(sin² 45 $°$ $-$ tan² 30 $°$ ) = 6

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Solution

$To show : 2 (\cos^{2} 45 ° + \tan^{2} 60 °) - 6 (\sin^{2} 45 ° - \tan^{2} 30 °) = 6 Here, LHS = 2 (\cos^{2} 45 ° + \tan^{2} 60 °) - 6 (\sin^{2} 45 ° - \tan^{2} 30 °) = 2 [{(\frac{1}{\sqrt{2}})}^{2} + {(\sqrt{3})}^{2}] - 6 [{(\frac{1}{\sqrt{2}})}^{2} - {(\frac{1}{\sqrt{3}})}^{2}] (From the table) = 2 (\frac{1}{2} + 3) - 6 (\frac{1}{2} - \frac{1}{3}) = 2 (\frac{1 + 6}{2}) - 6 (\frac{3 - 2}{6}) = 7 - 1 = 6 = RHS$

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