Question

Find sin(a)&cos(B), tan(a)&cot(B), and sec(a)&csc(B).​

Answer 1

Part A)
`sin(\alpha)=(4)/(7),\ cos(\beta)=(4)/(7)`

Part B)
`tan(\alpha)=(4)/(√(33)),\ tan(\beta)=(4)/(√(33))`

Part C)
`sec(\alpha)=(7)/(√(33)),\ csc(\beta)=(7)/(√(33))`

Explanation:

Part A) Find
`sin(\alpha)\ and\ cos(\beta)`

we know that

If two angles are complementary, then the value of sine of one angle is equal to the cosine of the other angle

In this problem

`\alpha+\beta=90^o` ---> by complementary angles

so

`sin(\alpha)=cos(\beta)`

Find the value of
`sin(\alpha)` in the right triangle of the figure

`sin(\alpha)=(8)/(14)` ---> opposite side divided by the hypotenuse

simplify

`sin(\alpha)=(4)/(7)`

therefore

`sin(\alpha)=(4)/(7)`

`cos(\beta)=(4)/(7)`

Part B) Find
`tan(\alpha)\ and\ cot(\beta)`

we know that

If two angles are complementary, then the value of tangent of one angle is equal to the cotangent of the other angle

In this problem

`\alpha+\beta=90^o` ---> by complementary angles

so

`tan(\alpha)=cot(\beta)`

Find the value of the length side adjacent to the angle alpha

Applying the Pythagorean Theorem

Let

x ----> length side adjacent to angle alpha

`14^2=x^2+8^2\\x^2=14^2-8^2\\x^2=132`

`x=√(132)\ units`

simplify

`x=2√(33)\ units`

Find the value of
`tan(\alpha)` in the right triangle of the figure

`tan(\alpha)=(8)/(2√(33))` ---> opposite side divided by the adjacent side angle alpha

simplify

`tan(\alpha)=(4)/(√(33))`

therefore

`tan(\alpha)=(4)/(√(33))`

`tan(\beta)=(4)/(√(33))`

Part C) Find
`sec(\alpha)\ and\ csc(\beta)`

we know that

If two angles are complementary, then the value of secant of one angle is equal to the cosecant of the other angle

In this problem

`\alpha+\beta=90^o` ---> by complementary angles

so

`sec(\alpha)=csc(\beta)`

Find the value of
`sec(\alpha)` in the right triangle of the figure

`sec(\alpha)=(1)/(cos(\alpha))`

Find the value of
`cos(\alpha)`

`cos(\alpha)=(2√(33))/(14)` ---> adjacent side divided by the hypotenuse

simplify

`cos(\alpha)=(√(33))/(7)`

therefore

`sec(\alpha)=(7)/(√(33))`

`csc(\beta)=(7)/(√(33))`