Question #b9f69

Question

1111 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-15T06:55:42

I hope that the explanation helps.

Given: $tan(x) = 5/12$ and x is acute.

Starting with the identity:

$sin^2(x) + cos^2(x) = 1$

Divide both sides by cos^2(x):

$sin^2(x)/cos^2(x) + 1 = 1/cos^2(x)$

Use the identity $sin(x)/cos(x) = tan(x)$ :

$tan^2(x) + 1 = 1/cos^2(x)$

Use the identity $1/cos(x) = sec(x)$

$tan^2(x) + 1 = sec^2(x)$

Substitute $(5/12)^2$ for $tan^2(x)$

$(5/12)^2 + 1 = sec^2(x)$

$25/144 + 1 = sec^2(x)$

$25/144 + 144/144 = sec^2(x)$

$169/144 = sec^2(x)$

$sec(x) = +-13/12$

But we drop the negative, because we are told that x is acute:

$sec(x) = 13/12$

$sec(x) = 1/cos(x)$

$cos(x) = 12/13$

$sin(x) = tan(x)cos(x)$

$sin(x) = 5/12 12/13$

$sin(x) = 5/13$

$csc(x) = 1/sin(x)$

$csc(x) = 13/5$

$cot(x) = 1/tan(x)$

$cot(x) = 12/5$