\(\displaystyle{\sin{{\left({A}+{B}\right)}}}\ne{\sin{{\left({A}\right)}}}+{\sin{{\left({B}\right)}}}\)

\(\displaystyle{\cos{{\left({A}+{B}\right)}}}\ne{\cos{{\left({A}\right)}}}+{\cos{{\left({B}\right)}}}\)

\(\displaystyle{\tan{{\left({A}+{B}\right)}}}\ne{\tan{{\left({A}\right)}}}+{\tan{{\left({B}\right)}}}\)

The statement is True.

Since the formulas for the trigonometric function of the sum of two angles are

\(\displaystyle{\sin{{\left({A}+{B}\right)}}}={\sin{{\left({A}\right)}}}{\cos{{\left({B}\right)}}}+{\cos{{\left({A}\right)}}}{\sin{{\left({B}\right)}}}\)

\(\displaystyle{\cos{{\left({A}+{B}\right)}}}={\cos{{\left({A}\right)}}}{\cos{{\left({B}\right)}}}+{\sin{{\left({A}\right)}}}{\sin{{\left({B}\right)}}}\)

\(\displaystyle{\tan{{\left({A}+{B}\right)}}}=\frac{{{\tan{{\left({A}\right)}}}+{\tan{{\left({B}\right)}}}}}{{{1}-{\tan{{\left({A}\right)}}}{\tan{{\left({B}\right)}}}}}\)

Hence generally the trigonometric function for the sum of two angles is not equal to the trigonometric function of the first angle plus trigonometry function of the second angle.