Question

tan A
write all the other trigonometric ratios of
in terms of Sec A​

Answer 1

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An equation involving trigonometric ratios of an angle is called a trigonometric identity. This is true for all angles A such that 0°≤ A ≤ 90°.

Some identities are:

1) cos² A + sin²A = 1

2) sec² A - tan² A = 1.

3) cosec² A - cot² A = 1.

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We know that,

sec A = 1 / cos A

⇒ cos A = 1/sec A

•cos²A + sin²A = 1

⇒  sin²A = 1 – cos²A

⇒  sin²A = 1 – (1/sec²A)                          (cosA= 1/secA)

⇒  sin²A = (sec²A-1)/sec²A

⇒sinA = √((sec²A-1)/sec²A)

⇒sinA = √(sec²A-1) ÷ (secA)_______(i)

•sin A = 1/cosec A

⇒ cosec A = 1/sin A

⇒cosecA= secA ÷√sec²A-1        (from eq i)

Now,

•sec²A – tan²A = 1

⇒ tan²A = sec²A + 1

⇒tanA = √sec²A + 1_______(ii)

•tan A = 1/cot A

⇒ cot A = 1/tan A

⇒cotA  = 1/√sec²A + 1        (from eq ii)

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Hope this will help you : )

Answer 2

Answer:

cos² A + sin²A = 1

2) sec² A - tan² A = 1.

3) cosec² A - cot² A = 1.

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We know that,

sec A = 1 / cos A

⇒ cos A = 1/sec A

•cos²A + sin²A = 1

⇒ sin²A = 1 – cos²A

⇒ sin²A = 1 – (1/sec²A) (cosA= 1/secA)

⇒ sin²A = (sec²A-1)/sec²A

⇒sinA = √((sec²A-1)/sec²A)

⇒sinA = √(sec²A-1) ÷ (secA)_______(i)

•sin A = 1/cosec A

⇒ cosec A = 1/sin A

⇒cosecA= secA ÷√sec²A-1 (from eq i)

Now,

•sec²A – tan²A = 1

⇒ tan²A = sec²A + 1

⇒tanA = √sec²A + 1_______(ii)

•tan A = 1/cot A

⇒ cot A = 1/tan A

⇒cotA = 1/√sec²A + 1 (from eq ii)