$\sin{(\theta+60^°)}$ $-$ $\cos{(\theta+30^°)}$ is a trigonometric expression and its value can be evaluated by simplifying it. The both trigonometric functions have angles which are sum of two angles so that the value of the trigonometric expression can be evaluated by using angle sum identities.
The trigonometric term $\sin{(\theta+60^°)}$ can be expanded by using sin of sum of two angles trigonometric identity.
$\sin{(\theta+60^°)}$ $\,=\,$ $\sin{\theta}\cos{(60^°)}$ $+$ $\sin{\theta}\cos{(60^°)}$
$\implies \sin{(\theta+60^°)}$ $\,=\,$ $\sin{\theta} \times \cos{(60^°)}$ $+$ $\cos{\theta} \times \sin{(60^°)}$
$\implies \sin{(\theta+60^°)}$ $\,=\,$ $\sin{\theta} \times \dfrac{1}{2}$ $+$ $\cos{\theta} \times \dfrac{\sqrt{3}}{2}$
$\,\,\, \therefore \,\,\,\,\,\, \sin{(\theta+60^°)}$ $\,=\,$ $\dfrac{\sin{\theta}}{2}$ $+$ $\dfrac{\sqrt{3}\cos{\theta}}{2}$
Similarly, the trigonometric term $\cos{(\theta+30^°)}$ can also be expanded by using cos of sum of two angles trigonometric identity.
$\cos{(\theta+30^°)}$ $\,=\,$ $\cos{\theta}\cos{(30^°)}$ $-$ $\sin{\theta}\sin{(30^°)}$
$\implies \cos{(\theta+30^°)}$ $\,=\,$ $\cos{\theta} \times \cos{(30^°)}$ $-$ $\sin{\theta} \times \sin{(30^°)}$
$\implies \cos{(\theta+30^°)}$ $\,=\,$ $\cos{\theta} \times \dfrac{\sqrt{3}}{2}$ $-$ $\sin{\theta} \times \dfrac{1}{2}$
$\,\,\, \therefore \,\,\,\,\,\, \cos{(\theta+30^°)}$ $\,=\,$ $\dfrac{\sqrt{3}\cos{\theta}}{2}$ $-$ $\dfrac{\sin{\theta}}{2}$
Finally, subtract $\cos{(\theta+30^°)}$ from $\sin{(\theta+60^°)}$ to get the value of the trigonometric expression mathematically.
$\sin{(\theta+60^°)}$ $-$ $\cos{(\theta+30^°)}$ $\,=\,$ $\Bigg(\dfrac{\sin{\theta}}{2}$ $+$ $\dfrac{\sqrt{3}\cos{\theta}}{2}\Bigg)$ $-$ $\Bigg(\dfrac{\sqrt{3}\cos{\theta}}{2}$ $-$ $\dfrac{\sin{\theta}}{2}\Bigg)$
$\implies \sin{(\theta+60^°)}$ $-$ $\cos{(\theta+30^°)}$ $\,=\,$ $\dfrac{\sin{\theta}}{2}$ $+$ $\dfrac{\sqrt{3}\cos{\theta}}{2}$ $-$ $\dfrac{\sqrt{3}\cos{\theta}}{2}$ $+$ $\dfrac{\sin{\theta}}{2}$
$\implies \sin{(\theta+60^°)}$ $-$ $\cos{(\theta+30^°)}$ $\,=\,$ $\dfrac{\sin{\theta}}{2}$ $+$ $\require{\cancel} \cancel{\dfrac{\sqrt{3}\cos{\theta}}{2}}$ $-$ $\require{\cancel} \cancel{\dfrac{\sqrt{3}\cos{\theta}}{2}}$ $+$ $\dfrac{\sin{\theta}}{2}$
$\implies \sin{(\theta+60^°)}$ $-$ $\cos{(\theta+30^°)}$ $\,=\,$ $\dfrac{\sin{\theta}}{2}$ $+$ $\dfrac{\sin{\theta}}{2}$
$\implies \sin{(\theta+60^°)}$ $-$ $\cos{(\theta+30^°)}$ $\,=\,$ $\dfrac{2\sin{\theta}}{2}$
$\implies \sin{(\theta+60^°)}$ $-$ $\cos{(\theta+30^°)}$ $\,=\,$ $\require{cancel} \dfrac{\cancel{2} \sin{\theta}}{\cancel{2}}$
$\,\,\, \therefore \,\,\,\,\,\, \sin{(\theta+60^°)}$ $-$ $\cos{(\theta+30^°)}$ $\,=\,$ $\sin{\theta}$
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