25.5

Qisqa javob:
a) \(1\); b) \(1\); d) \(1\); e) \(1\).
\(d\) va \(e\) da ifoda aniqlangan bo‘lishi kerak, ya’ni \(\alpha\neq 90^\circ k\).

Asosiy g‘oya. Keltirish formulalaridan foydalanamiz: \[ \cos(180^\circ-\alpha)=-\cos\alpha,\qquad \sin(180^\circ-\alpha)=\sin\alpha, \] \[ \cos(90^\circ-\alpha)=\sin\alpha,\qquad \sin(90^\circ-\alpha)=\cos\alpha, \] \[ \operatorname{tg}(90^\circ-\alpha)=\operatorname{ctg}\alpha,\qquad \operatorname{ctg}(90^\circ-\alpha)=\operatorname{tg}\alpha. \]

a) \[ \cos^2(180^\circ-\alpha)+\cos^2(90^\circ-\alpha) =(-\cos\alpha)^2+\sin^2\alpha =\cos^2\alpha+\sin^2\alpha=1. \]

b) \[ \sin^2(180^\circ-\alpha)+\sin^2(90^\circ-\alpha) =\sin^2\alpha+\cos^2\alpha=1. \]

d) \[ \operatorname{tg}\alpha\cdot\operatorname{tg}(90^\circ-\alpha) =\operatorname{tg}\alpha\cdot\operatorname{ctg}\alpha=1. \]

e) \[ \operatorname{ctg}\alpha\cdot\operatorname{ctg}(90^\circ-\alpha) =\operatorname{ctg}\alpha\cdot\operatorname{tg}\alpha=1. \]

Tekshirish: masalan, \(\alpha=30^\circ\) bo‘lsa, \[ \cos^2 150^\circ+\cos^2 60^\circ =\left(\frac{\sqrt3}{2}\right)^2+\left(\frac12\right)^2 =\frac34+\frac14=1. \] Demak, natijalar to‘g‘ri.

Eslab qolish uchun: avval \(180^\circ\pm\alpha\) yoki \(90^\circ\pm\alpha\) formulalarini qo‘llang, so‘ng \(\sin^2\alpha+\cos^2\alpha=1\) yoki \(\operatorname{tg}\alpha\cdot\operatorname{ctg}\alpha=1\) ayniyatlaridan foydalaning.