Question

which one of them is the external force of entrepreneurial motivation?
a. influence
b. strength
c. availability of resources
d.both A&C


give the answer​

Answer 1

Given:

Expression:$\frac{ \sin(30) + \tan(45 ) - \csc(60) }{ \sec(30) + \cos(60) + \cot(45) }$

To Find:

The value after the evaluation of the given expression

Solution:

★ The values here are,

sin(30) = 1/2

tan(45) = 1

csc(60) = 2/√3

sec(30) = 2/√3

cos(60) = 1/2

cot(45) = 1

★ Putting the values in the expression we get,

$\dashrightarrow \tt \: \frac{ \sin(30) + \tan(45) - \csc(60) }{ \sec(30) + \cos(60) + \cot(45) } \\ \\ \\ \dashrightarrow \tt \: \frac{ \cancel\dfrac{1}{2} + \cancel\dfrac{1}{1} + \cancel\dfrac{2}{ \sqrt{3} } }{\cancel \dfrac{2}{ \sqrt{3} } + \cancel\dfrac{1}{2} + \cancel \dfrac{1}{1} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ \dashrightarrow \tt \: \frac{1 + 1 + 1}{1 + 1 +1 } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ \dashrightarrow \tt \: \cancel\frac{3}{3} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ \dashrightarrow { \purple{ \underline{ \boxed{ \tt{ 1}}} \bigstar}} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

Hence:

The value of $\frac{ \sin(30) + \tan(45 ) - \csc(60) }{ \sec(30) + \cos(60) + \cot(45) }$ after evaluation results 1

More to know:

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

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Answer 2

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✧═════════════•❁❀❁•══════════════✧

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$\huge\star\huge{ \pink{\bold {\underline {\underline {\red {Great Day }}}}}} \star$

$\huge\blue\bigstar\pink{Be Safe Always }$

$\huge\fbox \red{✔Que} {\colorbox{crimson}{est}}\fbox\red{ion✔}$

The Question is Give below

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The Correct answer is option D

$\huge\color{cyan}\boxed{\colorbox{black}{Cute boy❤}}$

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