Solution :
Consider a triangle ABC, in which \(\angle\) B = 90
For \(\angle\) A, we have :
Base = AB,
Perp = BC
and Hyp = AC
\(\therefore\) sec A = \(Hyp\over Base\) = \(AC\over AB\)
So, \(AC\over AB\) = sec A = \(sec A\over 1\)
Let AB = k and AC = k sec A
So, BC = \(\sqrt{{AC}^2 – {AB}^2}\) = \(k\sqrt{sec^2 A – 1}\)
Now, sin A = \(BC\over AB\) = \(\sqrt{sec^2 A – 1}\over sec A\)
cos A = \(AB\over AC\) = \(1\over sec A\)
tan A = \(BC\over AB\) = \(\sqrt{sec^2 A – 1}\)
cot A = \(1\over tan A\) = \(1\over \sqrt{sec^2 A – 1}\)
cosec A = \(1\over sin A\) = \(sec A\over \sqrt{sec^2 A – 1}\)
