чему равно tg^2x+1/sinx+1/cosx+ctg^2x,если tgx+ctgx=5?. Created by keksonko. algebra-ru

cosx+ctg^2x,если tgx+ctgx=5?...

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чему равно tg^2x+1/sinx+1/cosx+ctg^2x,если tgx+ctgx=5?

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Eanhim $tg^2 x+\frac{1}{sinx}+\frac{1}{cosx}+ctg^2 x = \ ?, \ tgx+ctgx=5. \\
(tgx+ctgx)^2=25=tg^x \ x+ctg^2 x + 2 tg \ x \cdot ctg \ x = tg^x \ x+ctg^2 x + 2. \\
 tg^x \ x+ctg^2 x = 23, \ \frac{sinx}{cosx}+\frac{cosx}{sinx}=\frac {sin^2 x + cos^2 x}{sinx \cdot cosx} = \frac {1}{sinx \cdot cosx} =5. (I) \\
23+ \frac{cosx}{sinx \cdot cosx}+\frac{sinx}{sinx \cdot cosx} = 23+ \frac{1}{sinx \cdot cosx} \cdot (sinx + cosx) = \\$
$= 23+5 \cdot (sinx+cosx), \ \frac {1}{sinx \cdot cosx} =5 (I) \Rightarrow sinx \cdot cosx = \frac {1}{5} \\
(sinx+cosx)^2=sin^2 x + cos^2 x + 2 \cdot sinx \cdot cosx = 1 + \frac {2}{5} = \frac {12}{5} \\
sinx + cosx = 2\sqrt {\frac {3}{5}} \\
 23+5 \cdot (sinx+cosx) = 23+10\sqrt {\frac {3}{5}}$
Ответ: $tg^2 x+\frac{1}{sinx}+\frac{1}{cosx}+ctg^2 x = 23+10\sqrt {\frac {3}{5}}$

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abcd parallelogramm ae i df bissektrisy najti ugol mezhdu bissektrisami

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