Question

The value of 2(sin2 45°+ cot2 30°) – 6(cos2 45°— tan2 30°) is

Answer 1

To find:

The value of $2(sin^{2} 45 + cot^{2} 30 ) - 6(cos^{2} 45 - tan^{2} 30)$

Step-by-step explanation:

$2(sin^{2} 45 + cot^{2} 30 ) - 6(cos^{2} 45 - tan^{2} 30)$

= $2((\frac{1}{\sqrt{2} } )^{2} + (\sqrt{3}) ^{2} ) - 6 ((\frac{1}{\sqrt{2} } )^{2} - (\frac{1}{\sqrt{3} } )^{2})$

= $2(\frac{1}{2} + \frac{3}{1} ) - 6(\frac{1}{2} - \frac{1}{3} )$

= $2(\frac{7}{2} ) - 6(\frac{1}{6} )$

= $7-6$

= $1$

Answer:

Therefore, the value of $2(sin^{2} 45 + cot^{2} 30 ) - 6(cos^{2} 45 - tan^{2} 30)$ is $1$

Answer 2

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