Question

Решите
1+tg²β=ctgβsec²βsqrt(sec²β-1)

Answer 1

Ответ:

$0<\beta <\dfrac{\pi}{2}\\\\\\ctg\beta \cdot sec^2\beta \sqrt{sec^2\beta -1}=\dfrac{cos\beta }{sin\beta }\cdot \dfrac{1}{cos^2\beta }\cdot \sqrt{\dfrac{1}{cos^2\beta }-1}=\\\\\\=\dfrac{1}{sin\beta \cdot cos\beta }\cdot \sqrt{\dfrac{1-cos^2\beta }{cos^2\beta }}=\dfrac{1}{sin\beta \cdot cos\beta }\cdot \sqrt{\dfrac{sin^2\beta }{cos^2\beta }}=\dfrac{1}{sin\beta \cdot cos\beta }\cdot \dfrac{sin\beta }{cos\beta }=\\\\\\=\dfrac{1}{cos^2\beta }=1+tg^2\beta \\\\\\1+tg^2\beta =1+tg^2\beta$