Given \(sec \theta\) = \(13\over 12\), calculate all other trigonometric ratios.

Solution :

Consider a triangle ABC in which \(\angle\) A = \(\theta\) and \(\angle\) B = 90triangle

Then, Base = AB, perp = BC and Hypo. = AC,

\(\therefore\)  \(sec \theta\) = \(perp\over hypo\) = \(BC\over AC\) = \(3\over 4\)

Let AC = 13k and AB = 12k. Then,

By using Pythagoras Theorem,

\({BC}^2\) = \({AC}^2\) – \({AB}^2\)

= \(169k^2\) – \(125k^2\) = \(25k^2\)

\(\implies\) BC = 5k

\(sin \theta\) = \(BC\over AC\) = \(5\over 13\)

\(cos \theta\) = \(AB\over AC\) = \(12\over 13\)

\(tan \theta\) = \(BC\over AB\) = \(5\over 12\)

\(cot \theta\) = \(AB\over BC\) = \(12\over 5\)

\(cosec \theta\) = \(AC\over BC\) = \(13\over 5\)

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