Find the value of $4 {cos}^{2} 60^{0} + 4 {tan}^{2} 45^{0} - c o s e c^{2} 30^{0}$

0

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2

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1

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\frac{1}{2}

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Solution

The correct option is C $1$
Substituting the corresponding values we get
$4 {cos}^{2} 60^{0} + 4 {tan}^{2} 45^{0} - c o s e c^{2} 30^{0} = 4 {(\frac{1}{2})}^{2} + 4 (1)^{2} - 2^{2} = 1 + 4 - 4 = 1$

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Similar questions

4 {cos}^{2} 60^{0} + 4 {tan}^{2} 45^{0} - {csc}^{2} 30^{0}

\sqrt{\frac{1 - {sin}^{2} 30^{0}}{1 - {cos}^{2} 30^{0}}}

A = ⎡ ⎢ ⎣ \begin{matrix} 2 & 0 & 1 2 & 1 & 3 1 & - 1 & 0 \end{matrix} ⎤ ⎥ ⎦

A^{2} - 5 A + 6 I

x^{2} - 5 x - 1 = 0

x^{2} + \frac{1}{x^{2}}

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