Math, asked by beingdaring12, 2 months ago

find the value of 2 tan 45°+ cos30°- sin 60°​

Answers

Answered by 176646
1

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Answered by Anonymous
21

Given:

  • 2 tan 45°+ cos30°- sin 60°

 \\

To Find:

  • The value after evaluation

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Solution:

Values here are,

  • 2 tan 45° = 1
  • Cos 30° = √3/2
  • Sin 60° = √3/2

 \\

Putting the values in the equation we get,

 \dashrightarrow \tt \: 2 \tan(45)\degree  +   \cos(30)\degree  -  \sin(60 ) \degree \\  \\  \\ \\  \dashrightarrow \tt2(1) +  \frac{ \sqrt{3} }{2}  -  \frac{ \sqrt{3} }{2}  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \\  \\  \\  \\ \dashrightarrow \tt \: 2  \:  \:  \cancel{ + \frac{ \sqrt{3} }{2} } \:  \:  \:  \:  \:  \: \cancel{  -  \frac{ \sqrt{3} }{2} } \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:   \\  \\  \\ \\  \dashrightarrow { \purple{ \underline{ \boxed{ \pmb{ \frak{2}}}} \bigstar}} \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:

\\

Hence,

  • The value of 2 tan 45°+ cos30°- sin 60° is 2

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More to know:

\begin{gathered}\purple{\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}

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