Question

Find the value of 1-cos@/1+cos@, when @=90°​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Topic:-

Trigonometry

Question:-

$find \: the \: value \: of \: \dfrac{1 - cos \theta}{1 + cos \theta} \: when \: \theta \: = 90$

Solution:-

$\dfrac{1 - cos \theta}{1 + cos \theta}$

$we \: have \: to \: substitute \: the \: value \: of \: 90 \: in \: the \: place \: of \: \theta$

$= \dfrac{1 - cos90}{1 + cos90} \\ \\ = \dfrac{1 - 0}{1 + 0} \\ \\ = \dfrac{0}{0}$

So the value is not defined because anything divided by zero can't be determined

Answer:-

Can't be determined

More Information:-

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