Question

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

Answer 1

Answer:

LAST OPTION.

Step-by-step explanation:

Remember that:

$tan(x)=\frac{sin(x)}{cos(x)}$

$\frac{cos(x)}{sin(x)}=cot(x)=\frac{1}{tan(x)}$

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

$sin(90\°)=1\\cos(90\°)=0$

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

$tan(90-A)=\frac{sin(90-A)}{cos(90-A)}$

Applying the identities, you get:

$tan(90\°-A)=\frac{sin(90\°)cos(A)-cos(90\°)sin(A)}{cos(90\°)cos(A)-sin(90\°)sin(A))}$

Finally, you must simplify. Then:

$tan(90\°-A)=\frac{(1)cos(A)-(0)sin(A)}{(0)cos(A)-(1)sin(A))}\\\\tan(90\°-A)=\frac{cos(A)}{sin(A)}\\\\tan(90\°-A)=\frac{1}{tan(A)}$

Answer 2

Answer:

tan(90 - A) = 1/tan(A) ⇒ last answer

Step-by-step 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)