NEED Trig help?

Question

How do I solve this problem?
Express tan(b) in terms of x, given x=5csc(b) and 0<(b)<(pie)/2

Answer 1

x = 5csc(b)
csc(b)= x/5
0
tan(b)
= sin(b)/cos(b)
= sin(b)/(sqrt(1-sin^2(b)))
= (1/csc(b))/sqrt(1-1/csc^2(b))
= (5/x)/sqrt(1-25/x^2)
= 5/sqrt(x^2-25)

Answer 2

Express tan b in terms of x, that is x = 5csc b and 0
Or x/5 = csc b

csc b = sec b / tan b (Identity property)

Rearrange x/5 = sec b / tan b to 1/x = tan b / 5sec b and move 5sec b to the other side, so, 5sec b / x = tan b. Since, x = cscb, then tan b = 5sec b / csc b.

You can even simplify. Since sec b = 1/cos b and csc b is 1/sin b, then tan b = 5sin b / cos b. But the previous would do.

Answer 3

Draw a right triangle in the first quadrant of a circle so the b is the central angle. csc is 1/sin so x = 5(1/sin b) = 5/sin b

This means x sin b = 5, and sin b = 5/x

Sine is opposite over hypotenuse so the opposite is 5 and hyp. is x. Tangent is opposite over adjacent = 5/adj, so use Pythagorean Theorem to find the adjacent then enter it into 5/adj

Answer 4

x = 5/sin(b) => sin(b) = 5/x

cos(b) = sqrt(1-sin^2(b)) = sqrt(1 - 25/x^2) = +[sqrt(x^2 - 25)]/x

tan(b) = sin(b)/cos(b)

now just plug in the things we got for sin(b) and cos(b)



* sqrt means square root