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

Asked by Abhisek | 1 year ago |  62

##### Solution :-

Cos A function in terms of sec A:

sec A = $$\dfrac{1}{cos A}$$

⇒ cos A = $$\dfrac{1}{sec A}$$

sec A function in terms of sec A:

cos2A + sin2A = 1

Rearrange the terms

sin2A = 1 – cos2A

sin2A = 1 – ($$\dfrac{1}{sec^2 A}$$)

sin2A = $$\dfrac{(sec^2A-1)}{sec^2A}$$

sin A = ± $$\dfrac{\sqrt{(sec^{2}A-1}}{sec A}$$

cosec A function in terms of sec A:

sin A = $$\dfrac{1}{cosec A}$$

⇒cosec A = $$\dfrac{1}{sin A}$$

cosec A = ± $$\dfrac{sec A}{\sqrt{sec^2A-1}}$$

Now, tan A function in terms of sec A:

sec2A – tan2A = 1

Rearrange the terms

⇒ tan2A = sec2A – 1

tan A = $$( {\sqrt{sec^2A-1}})$$

cot A function in terms of sec A:

tan A = $$\dfrac{1}{cotA}$$

⇒ cot A = $$\dfrac{1}{tanA}$$

cot A = $$\dfrac{±1}{( {\sqrt{sec^2A-1}})}$$

Answered by Pragya Singh | 1 year ago

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