Question

Tan(π÷4+A) tan(3π÷4+A) =-1

Answer 1

Answer:

$\tan( \frac{\pi}{4} + \alpha ) \tan( \frac{3\pi}{4} + \alpha ) =$

$\frac{ \tan( \frac{\pi}{4} ) + \tan( \alpha ) }{1 - \tan( \frac{\pi}{4} ) \tan( \alpha ) } \times \frac{ \tan( \frac{3\pi}{4} ) + \tan( \alpha ) }{1 - \tan( \frac{3\pi}{4} ) \tan( \alpha ) } =$

$\frac{1 + \tan( \alpha ) }{1 - \tan( \alpha ) } \times \frac{ - 1 + \tan( \alpha ) }{1 + \tan( \alpha ) } = - 1$