The sine of angle seventy degrees is subtracted from the sine of angle fifty degrees, and sine of angle ten degrees is added to their difference.

$\sin{50^\circ}$ $-$ $\sin{70^\circ}$ $+$ $\sin{10^\circ}$

Firstly, let us check whether the sum of two angles of sine functions is a right angle or not. The sum of any two angles is not a right angle in this case. So, the cofunction identities cannot be used while evaluating the given trigonometric expression.

The difference of sines of angles can be transformed into product form by the difference to product identity of sine functions.

$=\,\,\,$ $2\cos{\bigg(\dfrac{50^\circ+70^\circ}{2}\bigg)}\sin{\bigg(\dfrac{50^\circ-70^\circ}{2}\bigg)}$ $+$ $\sin{10^\circ}$

Now, simplify the arithmetic expressions at angle position in each sine function for evaluating the trigonometric expression.

$=\,\,\,$ $2\cos{\bigg(\dfrac{120^\circ}{2}\bigg)}\sin{\bigg(\dfrac{-20^\circ}{2}\bigg)}$ $+$ $\sin{10^\circ}$

$=\,\,\,$ $2\cos{\bigg(\dfrac{\cancel{120^\circ}}{\cancel{2}}\bigg)}\sin{\bigg(\dfrac{\cancel{-20^\circ}}{\cancel{2}}\bigg)}$ $+$ $\sin{10^\circ}$

$=\,\,\,$ $2\cos{\big(60^\circ\big)}\sin{\big(-10^\circ\big)}$ $+$ $\sin{10^\circ}$

The trigonometric expression in trinomial form is reduced to binomial. Now, let us try to simplify this trigonometric expression further.

$=\,\,\,$ $2 \times \cos{\big(60^\circ\big)} \times \sin{\big(-10^\circ\big)}$ $+$ $\sin{10^\circ}$

The sine of angle ten degrees is not known to us but the value of cosine of sixty degrees is known to us. So, it can be substituted in the trigonometric expression for moving ahead in evaluating the trigonometric expression.

$=\,\,\,$ $2 \times \dfrac{1}{2} \times \sin{\big(-10^\circ\big)}$ $+$ $\sin{10^\circ}$

Now, it is time to focus on simplifying the trigonometric expression further.

$=\,\,\,$ $\dfrac{2 \times 1}{2} \times \sin{\big(-10^\circ\big)}$ $+$ $\sin{10^\circ}$

$=\,\,\,$ $\dfrac{2}{2} \times \sin{\big(-10^\circ\big)}$ $+$ $\sin{10^\circ}$

$=\,\,\,$ $\dfrac{\cancel{2}}{\cancel{2}} \times \sin{\big(-10^\circ\big)}$ $+$ $\sin{10^\circ}$

$=\,\,\,$ $1 \times \sin{\big(-10^\circ\big)}$ $+$ $\sin{10^\circ}$

$=\,\,\,$ $\sin{\big(-10^\circ\big)}$ $+$ $\sin{10^\circ}$

According to the sine even-odd trigonometric identity, the sine of negative ten degrees is equal to the negative of sine of angle ten degrees.

$=\,\,\,$ $-\sin{10^\circ}$ $+$ $\sin{10^\circ}$

$=\,\,\,$ $-\cancel{\sin{10^\circ}}$ $+$ $\cancel{\sin{10^\circ}}$

$=\,\,\, 0$

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