Question

evaluate the following
2cos²60⁰+4sec²30⁰-sin²45​

Answer 1

Topic:-

Trigonometry

Question:-

2cos²60°+4sec²30°-sin²45

Solution:-

2cos²60°+4sec²30°-sin²45°

$we \: know \: that \: \\ \\ \cos60 = \dfrac{1}{2} \\ \\ sec \: 30 = \dfrac{2}{ \sqrt{3} } \\ \\ sin \: 45 = 1$

$so \: substitute \: all \: formulas \: in \: the \: given \: question \:$

2cos²60°+4sec²30°-sin²45°

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

$= 2 \times \dfrac{1}{4} + \dfrac{4}{ 3} - 1$

$= 2 + \dfrac{4}{3} - 1 \\ \\ \\ = 2 - 1 + \dfrac{4}{3} \\ \\ \\ = 1 + \dfrac{4}{3} \\ \\ \\ = \dfrac{3 + 4}{3} \\ \\ \\ = \dfrac{7}{3}$

Answer:-

$\dfrac{7}{3}$

More Information:-

Trigon metric Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

csc²θ - cot²θ = 1

Trigometric relations

sinθ = 1/cscθ

cosθ = 1 /secθ

tanθ = 1/cotθ

tanθ = sinθ/cosθ

cotθ = cosθ/sinθ

Trigonmetric ratios

sinθ = opp/hyp

cosθ = adj/hyp

tanθ = opp/adj

cotθ = adj/opp

cscθ = hyp/opp

secθ = hyp/adj

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\sf Trigonometry\: Table \\ \begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \rm \infty \\ \\ \rm cosec A & \rm \infty & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm \infty \\ \\ \rm cot A & \rm \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0\end{array}}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$