Question

(sin a-cos a)в квадрате+(sin a +cos a)в квадрате                                                              (cos a *tg a)в квадрате+(sin a *tg a) в квадрате

Answer 1

1) $cos^2 alpha - 2 * sin alpha * cos alpha sin^2 alpha</var> <var>cos^2 alpha 2 * sin alpha * cos alpha sin^2 alpha =$  </var></p> <p><var></var> <img src=[/tex]cos^2 alpha + sin^2 alpha) = 2" title=" 2 * (cos^2 alpha + sin^2 alpha) = 2" alt=" 2 * (cos^2 alpha + sin^2 alpha) = 2" />

2) $<var>(cos alpha * sin alpha / cos alpha)^2 + (sin alpha</var> *<var> sin alpha / cos alpha)^2 = </var><var>$

 $sin^2 alpha + (sin^2 alpha/cos alpha)^2 = (sin^2 alpha*cos^2 alpha + sin^4 alpha)/cos^2 alpha =</var><var><var>$

$</var> (cos^2 alpha + sin^2 alpha)*sin^2 alpha/cos^2 alpha = </var><var>sin^2 alpha/cos^2 alpha = tg^2 alpha$  

 $2 * (<var>cos^2 alpha + sin^2 alpha</var>) = 2$

2) $<var>(cos alpha * sin alpha / cos alpha)^2 + (sin alpha</var> *<var> sin alpha / cos alpha)^2 = </var><var>$

 $sin^2 alpha + (sin^2 alpha/cos alpha)^2 = (sin^2 alpha*cos^2 alpha + sin^4 alpha)/cos^2 alpha =</var><var><var>$

$cos^2 alpha - 2 * sin alpha * cos alpha + sin^2 alpha</var> + <var>cos^2 alpha + 2 * sin alpha * cos alpha + sin^2 alpha =$  

 $2 * (<var>cos^2 alpha + sin^2 alpha</var>) = 2$

2) $<var>(cos alpha * sin alpha / cos alpha)^2 + (sin alpha</var> *<var> sin alpha / cos alpha)^2 = </var><var>$

 $sin^2 alpha + (sin^2 alpha/cos alpha)^2 = (sin^2 alpha*cos^2 alpha + sin^4 alpha)/cos^2 alpha =</var><var><var>$

[tex] (cos^2 alpha + sin^2 alpha)*sin^2 alpha/cos^2 alpha = sin^2 alpha/cos^2 alpha = tg^2 alpha" />

Хотел, как лучше, но он такой длинной строки не понимает