Logx(x+12)=2 решить логарифмическое уравнение

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Доказать тригонометрическое тождество

√( (1+cosα) / (1-cosα) )  - √( (1 - cosα) / (1+cosα) ) =2ctgα для  0 < α < π/2

решение :  * * * освобождение от иррациональности в знаменателе (числителе)  дроби  * * *

√( (1+cosα) / (1-cosα) )  - √( (1 - cosα) /(1+cosα) ) =

√( (1+cosα)²/ (1-cos²α) )  - √( (1 - cosα)² /(1- cos²α) ) =

√( (1+cosα)² /sin²α )  - √( (1 - cosα)² /sin²α )  =  

||  1+cosα  ≥ 0  для любого α ,  а  sinα > 0  т.к.  0 < α < π/2 ||

= (1+cosα) /sinα - (1 - cosα) /sinα = (1+cosα - 1 + cosα) /sinα =2cosα/sinα =

2ctgα .