Question

If sin teta= 1/2 then the vallu of sin teta +cosec teta is​

Answer 1

Answer:

5/2

Step-by-step explanation:

Given,

$sin\theta=\dfrac{1}{2}$

To Find :-

Value of :-

$sin\theta+cosec\theta$

Formula Required :-

$cosec\theta=\dfrac{1}{sin\theta}$

Solution :-

Method 1 :-

$cosec\theta=\dfrac{1}{\dfrac{1}{2}}$

$cosec\theta=2$

$sin\theta+cosec\theta=\dfrac{1}{2}+2$

$=\dfrac{1+4}{2}$

$=\dfrac{5}{2}$

Method 2 :-

$sin\theta=\dfrac{1}{2}$

$sin\theta=sin30^{\circ}$

Cancelling 'sin' on both sides :-

$\theta=30^{\circ}$

$\implies cosec\theta=cosec30^{\circ}$

$cosec30^{\circ}=2$

$sin\theta+cosec\theta=\dfrac{1}{2}+2$

$=\dfrac{1+4}{2}$

$=\dfrac{5}{2}$

Know More :-

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