Question

sina cosB-sin(a-B)​
cos a cos B - cos(a-B)
1-tgx / 1+tgx

Answer 1

Объяснение:

$sin \alpha cos \beta - sin( \alpha - \beta ) = sin \alpha cos \beta - sin \alpha cos \beta + cos \alpha sin \beta = cos \alpha sin \beta .$

$os \alpha cos \beta - cos( \alpha - \beta ) = cos \alpha cos \beta - cos \alpha os \beta - sin \alpha sin \beta = - sin \alpha sin \beta$

$\frac{1 - tgx}{1 + tgx} = \frac{1 - \frac{sinx}{cosx} }{1 + \frac{sinx}{cosx} } = \frac{cosx - sinx}{cosx + sinx} = \frac{sin( \frac{\pi}{2} - x) - sinx}{sin( \frac{\pi}{2} - x) + sinx } = \frac{2sin( \frac{\pi}{4} - x)cos \frac{\pi}{4} }{2sin \frac{\pi}{4}cos( \frac{\pi}{4} - x) } = tg( \frac{\pi}{4} - x)$