Question

(1+ tan@+sec @)(1+cot@-cosec @)​

Answer 1

$(1+tanA+secA)(1+cotA-cosecA)\\\\=(1+\frac{sinA}{cosA}+\frac{1}{cosA})(1+\frac{cosA}{sinA}-\frac{1}{sinA})\\ \\=(\frac{cosA+sinA+1}{cosA})(\frac{sinA+cosA-1}{sinA})$

Applying identity (a + b) (a - b) in [ (sin A + cos A) + 1 ) [ (sin A + cos A) - 1 ).

$=\frac{(sinA+cosA)^2-1^2}{sinA.cosA}\\ \\=\frac{sinA^2+cosA^2+2.sinA.cosA-1}{sinA.cosA}\\ \\=\frac{1+2.sinA.cosA-1}{sinA.cosA}\\ \\=\frac{2sinA.cosA}{sinA.cosA}\\ \\=2$

Hope it helps.....   :D

# HaRd_WoRk...