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Mathematics

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

Trigonometric Identities

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Answer

Converting sin A in terms of cot A,

⇒sin A⇒sin2A⇒sin2A1⇒sin2Asin2A+ cos2A⇒1sin2Asin2A+cos2Asin2A⇒11+cot2A.\Rightarrow \text{sin A} \\[1em] \Rightarrow \sqrt{\text{sin}^2 A} \\[1em] \Rightarrow \dfrac{\sqrt{\text{sin}^2 A}}{1} \\[1em] \Rightarrow \dfrac{\sqrt{\text{sin}^2 A}}{\sqrt{\text{sin}^2 A + \text{ cos}^2 A}} \\[1em] \Rightarrow \dfrac{1}{\sqrt{\dfrac{\text{sin}^2 A}{\text{sin}^2 A} + \dfrac{\text{cos}^2 A}{\text{sin}^2 A}}} \\[1em] \Rightarrow \dfrac{1}{\sqrt{1 + \text{cot}^2 A}}.

Converting sec A in terms of cot A,

We know that,

⇒sec2A=1 + tan2A⇒sec2A=1+1cot2A⇒sec2A=cot2A+1cot2A⇒sec A=cot2A+1cot2A⇒sec A=1+cot2Acot A.\Rightarrow \text{sec}^2 A = \text{1 + tan}^2 A \\[1em] \Rightarrow \text{sec}^2 A = 1 + \dfrac{1}{\text{cot}^2 A}\\[1em] \Rightarrow \text{sec}^2 A = \dfrac{\text{cot}^2 A + 1}{\text{cot}^2 A} \\[1em] \Rightarrow \text{sec A} = \sqrt{\dfrac{\text{cot}^2 A + 1}{\text{cot}^2 A}} \\[1em] \Rightarrow \text{sec A} = \dfrac{\sqrt{1 + \text{cot}^2 A}}{\text{cot A}}.

We know that,

tan A = 1cot A\dfrac{1}{\text{cot A}}

Hence, sin A = 11+cot2A,sec A=1+cot2Acot A and tan A=1cot A.\dfrac{1}{\sqrt{1 + \text{cot}^2 A}}, \text{sec A} = \dfrac{\sqrt{1 + \text{cot}^2 A}}{\text{cot A}} \text{ and tan A} = \dfrac{1}{\text{cot A}}.

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