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)
Answer:
LAST OPTION.
Step-by-step explanation:
Remember that:
[tex]tan(x)=\frac{sin(x)}{cos(x)}[/tex]
[tex]\frac{cos(x)}{sin(x)}=cot(x)=\frac{1}{tan(x)}[/tex]
[tex]sin(x\±y) = sin(x)cos(y)\±cos(x)sin(y)\\\\cos(x\±y) =cos(x)cos(y)\± sin(x)sin (y) [/tex]
[tex]sin(90\°)=1\\cos(90\°)=0[/tex]
Then, you need to rewrite [tex]tan(90-A)[/tex]:
[tex]tan(90-A)=\frac{sin(90-A)}{cos(90-A)}[/tex]
Applying the identities, you get:
[tex]tan(90\°-A)=\frac{sin(90\°)cos(A)-cos(90\°)sin(A)}{cos(90\°)cos(A)-sin(90\°)sin(A))}[/tex]
Finally, you must simplify. Then:
[tex]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)}[/tex]
