# How do I find tan(𝜃) + sin(𝜃) for the angle 𝜃?.? Relevance
• The length of the hypotenuse

h=sqr(9^2+5^2)=sqr(106)

=>

tan(theta)+sin(theta)

=

5/9+5/sqr(106)

=

1.041198487

approximately.

• tan(𝜃) + sin(𝜃)

= 5/9 + 5/sqrt 106

= 1.0411984867341877

• tan(t) + sin(t) =>

tan(t) + 1/csc(t) =>

tan(t) + 1/sqrt(csc(t)^2) =>

tan(t) + 1/sqrt(1 + cot(t)^2) =>

(5/9) + 1/sqrt(1 + (9/5)^2) =>

(5/9) + 1/sqrt(1 + 81/25) =>

(5/9) + 1/sqrt((25 + 81)/25) =>

(5/9) + sqrt(25/(25 + 81)) =>

(5/9) + sqrt(25/106) =>

(5/9) + 5/sqrt(106) =>

(5/9) + 5 * sqrt(106) / 106 =>

(5 * 106 + 5 * 9 * sqrt(106)) / (9 * 106) =>

5 * (106 + 9 * sqrt(106)) / 954 =>

5 * (9 + sqrt(106)) * sqrt(106) / 954

• Remember the SOH-CAH-TOA mnemonic:

SOH --> Sin(θ) = Opposite/Hypotenuse

In your picture you have the Opposite side (5) and the Adjacent side (9), so tan(θ) is really easy.

tan(θ) = 5/9

For sin(θ) you need the hypotenuse. Do you remember the Pythagorean Theorem?

a² + o² = h²

9² + 5² = h²

81 + 25 = h²

106 = h²

h = √106

Now can *you* find an expression for sin(θ)? Add them together.