Question

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

Answer 1

Step-by-step explanation:

sin 60=cos 30

LHS =sin 60 X cos 30- cos 30 X sin 60

= sin ²60 - sin ²60

= 0

question is incorrect the sin in between should be (+)

Answer 2

Topic:-

Trigonometry

Question:-

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

Solution:-

$here \: are \: the \: values \: of \: \sin60 \: and \: cos60$

$\sin60 = \frac{ \sqrt{3} }{2} \\ \cos60 = \frac{ \sqrt{3} }{2} \\ \\ so \\ sin60 = cos30$

now substitute all formulas in the given question

$\frac{ \sqrt{3} }{2} \times \frac{ \sqrt{3} }{2} - \frac{ \sqrt{3} }{2} \times \frac{ \sqrt{3} }{2}$

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

$= \frac{3}{4} - \frac{3}{4}$

We Know That

anything number subtracted from itself =0

So, Answer=0

Answer:-

0

Know More:-

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