Question

1+cosecA / cosecA = cos^2 A / 1 - sinA​

Answer 1

Answer:

Step-by-step explanation:

To Prove :-

$\dfrac{1+cosecA}{cosecA}=\dfrac{cos^2A}{1-sinA}$

Solution :-

Taking L.H.S :-

$=\dfrac{1+cscA}{cscA}$

$=\dfrac{1+\dfrac{1}{sinA}}{\dfrac{1}{sinA}}$

$=\dfrac{\dfrac{sinA+1}{sinA}}{\dfrac{1}{sinA}}$

$= sinA + 1$

$= 1 + sinA\times \dfrac{1-sinA}{1-sinA}$

$=\dfrac{1-sin^2A}{1-sinA}$

$=\dfrac{cos^2A}{1-sinA}$

= R.H.S

Know More :-

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