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Q.1 Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.
Answer :


Consider a \( ∆ \ ABC \)

\( cotA \ = \ \frac{B}{P} \ = \ \frac{AB}{BC} \

=> \ \frac{AB}{BC} \ = \ \frac{cotA}{1} \)

Let \(AB = kcotA\) and \(BC = k\).

By Pythagoras Theorem,

\( AC \ = \ \sqrt{AB^2 + BC^2} \ = \ \sqrt{k^2cot^2A + k^2} \ = \ k \sqrt{1 + cot^2A} \)



∴ \( sinA \ = \ \frac{P}{B} \ = \ \frac{BC}{AC} \ = \ \frac{k}{k \sqrt{1+cot^2A}} \ = \ \frac{1}{ \sqrt{1+cot^2A}} \)

\( secA \ = \ \frac{H}{B} \ = \ \frac{AC}{AB} \ = \ \frac{k \sqrt{1+cot^2A}}{kcotA} \ = \ \frac{ \sqrt{1+cot^2A}}{cotA} \)

and, \( tanA \ = \ \frac{P}{B} \ = \ \frac{BC}{AB} \ = \ \frac{k}{kcotA} \ = \ \frac{1}{cotA} \)