Question

cos 60+ sin 150+ Cosec 120​

Answer 1

Answer:

cos 60°+ sin 150°+ Cosec 120°= 3

Step-by-step explanation:

cos 60°+ sin 150°+ Cosec 120°

=1 /2 + sin (90°+60°) + cosec (90°+30°)

=1/2 + cos 60° + sec 30°

=1/2+1/2+2

=3

so, cos 60° + sin 150° + Cosec 120° = 3

Answer 2

Given to find the value of

cos60° + sin150° + csc 120°

SOLUTION:-

Already we know that

cos60° = 1/2

sin150° It can be written as

sin150° = sin(90°+60°)

As we know that sin(90°+θ) = cosθ

So,

sin150° = cos60°

sin150° = 1/2

cosec120° = cosec(90°+30°)

As we know that cosec(90°+θ) = secθ Since ,

cosec120° = sec30°

cosec120° = 2/√3

So,

cos 60°+ sin 150°+ Cosec 120°

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

$\dfrac{ \sqrt{3} + \sqrt{3} + 4 }{2 \sqrt{3} }$

$\dfrac{2 \sqrt{3} + 4 }{2 \sqrt{3} }$

$\dfrac{2( \sqrt{3} + 2)}{2 \sqrt{3} }$

$\dfrac{\sqrt{3}+2}{\sqrt{3}}$

Know more :-

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