Let's calculate the value of the expression under a) and simplify the expression under b) as much as possible;
a) $\left(1-\cos 15^{\circ}\right)\left(1+\sin 75^{\circ}\right)+\cos 75^{\circ} \cos 15^{\circ} \operatorname{cotg} 15^{\circ}$
b) $\sin \left(45^{\circ}-\alpha\right)-\cos \left(30^{\circ}+\alpha\right)+\sin ^{2} 30^{\circ}-\cos \left(45^{\circ}+\alpha\right)+$ $+\sin \left(60^{\circ}-\alpha\right)+\sin ^{2} 60^{\circ}$.
Sure, here is the translation:

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When calculating the two expressions, we use the relationships between the trigonometric functions of complementary angles.

$$
\text { a) } \begin{gathered}
\left(1-\cos 15^{\circ}\right)\left(1+\sin 75^{\circ}\right)+\cos 75^{\circ} \cos 15^{\circ} \operatorname{cotg} 15^{\circ}= \\
=\left(1-\cos 15^{\circ}\right)\left(1+\cos 15^{\circ}\right)+\sin 15^{\circ} \cos 15^{\circ} \frac{\cos 15^{\circ}}{\sin 15^{\circ}}= \\
=1-\cos ^{2} 15^{\circ}+\cos ^{2} 15^{\circ}=1
\end{gathered}
$$

b) Notice that the complementary angle of \(30^{\circ}+\alpha\) is \(60^{\circ}-\alpha\). Therefore,

$$
\begin{aligned}
& \sin \left(45^{\circ}-\alpha\right)-\cos \left(30^{\circ}-\alpha\right)+\sin ^{2} 30^{\circ}-\cos \left(45^{\circ}+\alpha\right)+\sin \left(60^{\circ}-\alpha\right)+ \\
& +\sin ^{2} 60^{\circ}=\sin \left(45^{\circ}-\alpha\right)-\cos \left(30^{\circ}+\alpha\right)+\sin ^{2} 30^{\circ}-\sin \left(45^{\circ}-\alpha\right)+ \\
& +\cos \left(30^{\circ}+\alpha\right)+\cos ^{2} 30^{\circ}=\sin ^{2} 30^{\circ}+\cos ^{2} 30^{\circ}=1
\end{aligned}
$$

Eva and Maria Beke (Bp., XIII., Civil Engineering Technology II.o.t.)

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