Question

please solve this ........if you know then answer ​

Answer 1

Step-by-step explanation:

i hope it helps you

see only second image

sorry for late reply

Answer 2

$\tt \longrightarrow \dfrac{cos \: 45 \degree}{sec \: 30\degree + cosec \: 30\degree }$

We know that :-

$\tt cos \: 45 \degree = \dfrac{1}{ \sqrt{2} } \\ \\ \tt sec \: 30\degree = \dfrac{2}{ \sqrt{3} } \\ \\ \tt cosec \: 30\degree = 2$

$\tt \longrightarrow \dfrac{ \dfrac{1}{ \sqrt{2} } }{ \frac{2}{ \sqrt{3} } + 2 }$

$\tt \longrightarrow \dfrac{ \dfrac{1}{ \sqrt{2} } }{ \dfrac{2 + 2\sqrt{3}}{ \sqrt{3} } }$

$\tt \longrightarrow \dfrac{\sqrt{3}}{\sqrt{2}(2 + 2\sqrt{3})} = \dfrac{\sqrt{3}}{2\sqrt{2}(1 + \sqrt{3})}$

Answer :

$\tt \longrightarrow \dfrac{\sqrt{3}}{2\sqrt{2}(1 + \sqrt{3})}$

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