Question

sin 60° × cos 30° - cos 30° × sin60°= sin 30°​

Answer 1

Answer:

see the attachment. It is helpful to you.

Step-by-step explanation:

LHS not equal to RHS

Answer 2

$\textbf{\large{\underline{\pink{ Answer: }}}}$

Trigonometry

Question:-

sin 60° × cos 30° - cos 30° × sin60°= sin 30°

Solution:-

$\sin60 = \frac{ \sqrt{3} }{2} \\ \cos30 = \frac{ \sqrt{3} }{2} \\ now \: we \: have \: to \: substitute \: these \: values \: in \: the \: given \: question$

$\frac{ \sqrt{3} }{2} \times \frac{ \sqrt{3} }{2} - \frac{ \sqrt{3} }{2} \times \frac{ \sqrt{3} }{2} \\ = ( \frac{ \sqrt{3} }{2} {)}^{2} - ( \frac{ \sqrt{3} }{2} {)}^{2} \\ = \frac{3}{4} - \frac{3}{4} \\ = 0$

Answer:-

0

Know More:-

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