Question

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

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Answer 1

Cos A function in terms of sec A:

sec A = 1/cos A

⇒ cos A = 1/sec

sec A function in terms of sec A:

cos2A + sin2A = 1

Rearrange the terms

sin2A = 1 – cos2A

sin2A = 1 – (1/sec2A)

sin2A = (sec2A-1)/sec2A

sin A = ± √(sec2A-1)/sec A

cosec A function in terms of sec A:

sin A = 1/cosec A

⇒cosec A = 1/sin A

cosec A = ± sec A/√(sec2A-1)

Now, tan A function in terms of sec A:

sec2A – tan2A = 1

Rearrange the terms

⇒ tan2A = sec2A + 1

tan A = √(sec2A + 1)

cot A function in terms of sec A:

tan A = 1/cot A

⇒ cot A = 1/tan A

cot A = ±1/√(sec2A + 1)

Answer 2

Step-by-step explanation:

We know,

cosA=

secA

1

We know,

sin

2

A+cos

2

A=1

sin

2

A=1−cos

2

A

sinA=

1−(

secA

1

)

2

=

sec

2

A

sec

2

A−1

∴sinA=

sec

2

A

sec

2

A−1

We know,

cosec A=

sinA

1

∴cosec A=

sec

2

A−1

secA

We know,

cotA=

sinA

cosA

=

sinA

cosA

=

secA

sec

2

A−1

secA

1

∴cotA=

sec

2

A−1

1

We know,

tanA=

cotA

1

∴tanA=

sec

2

A−1