Question

express the trigonometric ratios sin A, sec A and Tan A in terms of cot A​

Answer 1

We have

1]cosec

2

A−cot

2

A=1

⇒cosec

2

A=1+cot

2

A

⇒(cosecA)

2

=cot

2

A+1

⇒(

sinA

1

)

2

=cot

2

+1

⇒(sinA)

2

=

cot

2

A+1

1

⇒sinA=±

cot

2

A+1

1

We reject the negative value of sinA for acute angle A. Therefore sinA=

cot

2

A+1

1

2]tanA=

cotA

1

3]We have sec

2

A−tan

2

A=1

⇒sec

2

A=1+tan

2

A

⇒sec

2

A=1+

cot

2

A

1

=

cot

2

A

cot

2

A+1

⇒secA=

cotA

cot

2

A+1

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Answer 2

(1) We know that  

cosec^2A − cot^2A = 1

⇒ cosec^2A = 1+ cot^2A

⇒cosecA= 1 ÷ (√1+cot^2A)

⇒ sinA =1 ÷ (√1+cot2A)            

{ °•° sinA=1 ÷ cosecA }

_______________________

(2) We know that  

sec^2A −tan^2A = 1

⇒sec^2A=1+tan^2A

⇒sec^2A =1+1 ÷ cot^2A

=cot^2A+1 ÷ cot^2A

⇒secA= (√1+cot^2A) ÷ cotA

_______________________

(3) We know that tanA=1 ÷ cotA

_______________________