Question

Express the trigonometric ratio of sin A and sec A and tan A in terms of cot A.

Answer 1

tan A

We know that

tan A = \(\frac{1}{cot A}\)

cosec A

We know that

1 + cot² A = cosec² A

cosec² A = 1 + cot² A

cosec A = \(\pm \sqrt{1 + cot^2 A}\)

Here, A is acute angle (i.e. less than 90°) & cosec A is positive when A is acute

cosec A = \(\sqrt{1 + cot^2 A}\)

sin A

sin A = \(\frac{1}{cosec A}\)

Putting value of cosec A found above

= \(\frac{1}{\sqrt{1 + cot^2 A}}\)

sec A

We know that

1 + tan² A = sec² A

sec² A = 1 + tan² A

sec A = \(\pm \sqrt{(1 + tan^2 A)}\)

Here, A is acute angle (i.e. less than 90°) & sec A is positive when A is acute

\(sec A = \sqrt{(1 + tan^2 A)}\)

Converting tan A to cot A

\(= \sqrt{(1 + \frac{1}{cot^2}})\)

\(= \sqrt{( \frac{cot^2 A + 1}{cot^2 A})}\)

\(= \frac{\sqrt {cot^2 A + 1}}{cot A}\)