Question

Write all the other trigonometric ratios of A in terms of

secA.

Answer 1

Solution:

We know that sec2A−tan2A=1

 

⇒tan2A=sec2A−1

 

⇒tanA=sec2A−1−−−−−−−−√     (1)

 

 

We know that cos A = 1secA      (2)

 

 

We know that sin2A+cos2A=1

⇒sin2A=1−cos2A

 

 

Putting (2) in the above equation we get

 

sin2A=1−(1secA)2=sec2A−1sec2A

 

⇒sinA=sec2A−1√secA                        (3)

 

 

We know that  cosecA=1sinA

 

 

Putting (3) in the above equation, we get

 

cosecA=secAsec2A−1√         (4)

 

 

We know that  cotA=1tanA

 

Putting (1) in the above equation, we get

 

cotA=1sec2A−1√                    (5)

 

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

Answer 2

$\huge\underline\mathfrak\color {red}Solution:$

We know that sec2A−tan2A=1

⇒tan2A=sec2A−1

⇒tanA=sec2A−1−−−−−−−−√     (1)

We know that cos A = 1secA      (2)

We know that sin2A+cos2A=1

⇒sin2A=1−cos2A

Putting (2) in the above equation we get

sin2A=1−(1secA)2=sec2A−1sec2A

⇒sinA=sec2A−1√secA                        (3)

We know that  cosecA=1sinA

Putting (3) in the above equation, we get

cosecA=secAsec2A−1√         (4)

We know that  cotA=1tanA

Putting (1) in the above equation, we get

cotA=1sec2A−1√                    (5)

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