Question

tan tita+sec tita-1/ tan tita -sec tita+1=1+sin tita /cos tita prove that

Answer 1

Trigonometry question...

 $\frac{tan\phi+sec\phi-1}{tan\phi-sec\phi+1}=\frac{tan\phi+sec\phi-1}{tan\phi-sec\phi+1}*\frac{tan\phi+sec\phi-1}{tan\phi+sec\phi-1}\\\\=\frac{tan^2\phi+sec^2\phi+1+2sec\phi tan\phi-2sec\phi-2tan\phi}{tan^2\phi-(sec^2\phi+1-2sec\phi)}\\\\=\frac{2sec^2\phi+2sec\phi tan\phi-2sec\phi-2tan\phi}{2(-1+sec\phi)}\\\\=\frac{1+sin\phi-cos\phi-sin\phi cos\phi}{ cos\phi(-cos\phi+1)}\\\\=\frac{(-cos\phi+1)(1+sin\phi)}{cos\phi(-cos\phi+1)}\\\\=\frac{1+sin\phi}{cos\phi}$