Question

TanX= sinX/cosX. Therefore, tan(90-A)= . (All angle measurements are in degrees.)

1/tan(90-A)
1/sin A
1/cos(90-A)
1/tan A

Answer 1

LAST OPTION.

Explanation:

Remember that:

`tan(x)=(sin(x))/(cos(x))`

`(cos(x))/(sin(x))=cot(x)=(1)/(tan(x))`

`sin(x\±y) = sin(x)cos(y)\±cos(x)sin(y)\\\\cos(x\±y) =cos(x)cos(y)\± sin(x)sin (y)`

`sin(90\°)=1\\cos(90\°)=0`

Then, you need to rewrite
`tan(90-A)`:

`tan(90-A)=(sin(90-A))/(cos(90-A))`

Applying the identities, you get:

`tan(90\°-A)=(sin(90\°)cos(A)-cos(90\°)sin(A))/(cos(90\°)cos(A)-sin(90\°)sin(A)))`

Finally, you must simplify. Then:

`tan(90\°-A)=((1)cos(A)-(0)sin(A))/((0)cos(A)-(1)sin(A)))\\\\tan(90\°-A)=(cos(A))/(sin(A))\\\\tan(90\°-A)=(1)/(tan(A))`

Answer 2

tan(90 - A) = 1/tan(A) ⇒ last answer

Explanation:

* Lets revise some important information for the right triangle

- In any right triangle

# The side opposite to the right angle is called the hypotenuse

# The other two sides are called the legs of the right angle

* If the name of the triangle is ABC, where B is the right angle

∴ The hypotenuse is AC

∴ AB and BC are the legs of the right angle

- ∠A and ∠C are two acute angles

∵ The sum of the interior angles of any triangle is 180°

∵ m∠B = 90°

∴ m∠A + m∠C = 180° - 90° = 90°

- If the measure of angle A is x°

∴ The measure of angle C = 90° - x°

- tan(A) = opposite/adjacent

∵ The opposite to ∠A is BC

∵ The adjacent to ∠A is AB

∴ tan(A) = BC/AB ⇒(1)

- tan(C) = opposite/adjacent

∵ The opposite to ∠C is AB

∵ The adjacent to ∠C is BC

∴ tan(C) = AB/BC ⇒ (2)

- From (1) and (2)

∴ tan(A) = 1/tan(C)

∵ m∠A = A , m∠C = (90 - A)

∴ tan(A) = 1/tan(90 - A)

OR

∴ tan(90 - A) = 1/tan(A)