Write the other trigonometric ratios of A in terms of sec A.

Solution :

Consider a triangle ABC, in which \(\angle\) B = 90triangle

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 = \(1\over sin A\) = \(sec A\over \sqrt{sec^2 A – 1}\)

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