Question

write all the trigonometric ratios in angle in terms of secA​

Answer 1

Answer:

Step-by-step explanation:

We know that  

sec^2A−tan^2A=1

 ⇒tan^2A=sec^2A−1

 ⇒tanA= √(sec2A−1) .......... eq (1)

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We know that  

cos A = 1 ÷ secA      .......... eq (2)

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 We know that  

sin^2A + cos^2A =1

⇒sin^2A = 1 − cos^2A

Putting (2) in the above equation we get

sin2A=1−(1÷ secA)^2 = (sec^2A−1)÷ sec^2A

⇒sinA= √(sec2A−1) ÷ secA           .......eq  (3)

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We know that  

cosecA= 1 ÷ sinA

Putting (3) in the above equation, we get

cosecA = secA ÷ (√sec^2A−1)     .......eq (4)

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We know that  

cotA = 1 ÷ tanA

Putting (1) in the above equation, we get

cotA= 1 ÷ (√sec2A−1)          ........ eq (5)

 (1), (2), (3), (4) and (5) shows all the other trigonometric ratios of ∠A in terms of sec A.

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