Question

prove that root of 1-cos o÷1+coso=coseco-coto​

Answer 1

We have to find out the value of :-

$\rm \longrightarrow \cos(1^{ \circ} ) \times \cos(2 ^ {\circ}) \times ... \cos(90^{ \circ} )⟶cos$

We know that :-

Value of cos 90° = 0

$\rm \longrightarrow \cos(1^{ \circ} ) \times \cos(2 ^ {\circ}) \times ... \cos(89^{ \circ} )\times 0$

$\rm \longrightarrow 0$

Answer :

$\rm \cos(1^{ \circ} ) \times \cos(2 ^ {\circ}) \times ... \cos(90^{ \circ} )$

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Trigonometric Table :-

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