1) (1+ ctg B)^2 + (1 - ctg B )^2
2)tg x + cos x/1 + sin x
3)cos B/1 - sin b + 1-sin b/cos B
4) 1 + tg a/1+ ctg a

Question

Алгебра

Nualymav

6 год назад

1) (1+ ctg B)^2 + (1 - ctg B )^2
2)tg x + cos x/1 + sin x
3)cos B/1 - sin b + 1-sin b/cos B
4) 1 + tg a/1+ ctg a

ОТВЕТЫ

Феофан

Jul 1, 2019

1) (1+ ctgβ)²+ (1 - ctgβ )²=1+2ctgβ+ctg²β+1-2ctgβ+ctg²β=
=2+2ctg²β=2(1+ctg²β)=1/sin²β;
2) $tgx+ \frac{cosx}{1+sinx} = \frac{tgx(1+sinx)+cosx}{1+sinx} = \frac{tgx+ \frac{sin ^{2}x }{cosx}+cosx }{1+sinx}= \frac{sinx+sin ^{2}x+cos^{2}x}{cosx}* \\ * \frac{1}{1+sinx}= \frac{1+sinx}{cosx} * \frac{1}{1+sinx}= \frac{1}{cosx}; \\$
3) $\frac{cos \beta }{1-sin \beta}+ \frac{1-sin \beta }{cos \beta }= \frac{cos^{2} \beta +1-2sin \beta +sin^{2} \beta }{(1-sin \beta )cos \beta } = \frac{2(1-sin \beta )}{(1-sin \beta )cos \beta }= \frac{2}{cos \beta };$
4) $\frac{1+tg \alpha }{1+ctg \alpha } = \frac{cos \alpha+sin \alpha }{cos \alpha } * \frac{sin \alpha }{sin \alpha +cos \alpha }= \frac{sin \alpha }{cos \alpha } =tg \alpha .$

159

1) (1+ ctg B)^2 + (1 - ctg B )^2 2)tg x + cos x/1 + sin x 3)cos B/1 - sin b + 1-sin b/cos B 4) 1 + tg a/1+ ctg a

1) (1+ ctg B)^2 + (1 - ctg B )^2
2)tg x + cos x/1 + sin x
3)cos B/1 - sin b + 1-sin b/cos B
4) 1 + tg a/1+ ctg a