Question

If cot theta = √9-x² , find sec theta and tan theta​

Answer 1

Step-by-step explanation:

Hypotenuse : BC = 13

Adjacent Side : AB = 13 (Adjacent to ∠ Θ)

Opposite Side : AC = ? (Opposite to ∠ Θ)

To find the Opposite Side - AC , we use Pythagoras theorem ((Hypotenuse)2 = (Leg 1)2 + (Leg 2)2)

____________

AC = √ BC2 - AB2

____________

AC = √ 132 - 112

____________

AC = √ 169 - 121

__

AC = √ 48

Now, we will simplify square root

_________________

AC = √ 2 x 2 x 2 x 2 x 3

__

AC = 2 + 2 √ 3 (pairs taken out)

__

AC = 4 √ 3

So Cosec, Sec and Cot will be as follows -

Cosec Θ = Hypotenuse / Opposite Side

__

Sin Θ = 13 / 4√ 3

Cot Θ = Adjacent Side / Opposite Side

__

Tan Θ = 11 / 4√ 3

Sec Θ = Hypotenuse / Adjacent Side

Sec Θ = 13/11

Answer 2

Step-by-step explanation:

Hypotenuse : BC = 13

Adjacent Side : AB = 13 (Adjacent to ∠ Θ)

Opposite Side : AC = ? (Opposite to ∠ Θ)

To find the Opposite Side - AC , we use Pythagoras theorem ((Hypotenuse)2 = (Leg 1)2 + (Leg 2)2)

____________

AC = √ BC2 - AB2

____________

AC = √ 132 - 112

____________

AC = √ 169 - 121

__

AC = √ 48

Now, we will simplify square root

_________________

AC = √ 2 x 2 x 2 x 2 x 3

__

AC = 2 + 2 √ 3 (pairs taken out)

__

AC = 4 √ 3

So Cosec, Sec and Cot will be as follows -

Cosec Θ = Hypotenuse / Opposite Side

__

Sin Θ = 13 / 4√ 3

Cot Θ = Adjacent Side / Opposite Side

__

Tan Θ = 11 / 4√ 3

Sec Θ = Hypotenuse / Adjacent Side

Sec Θ = 13/11