Question

tan theta= -3/4, angle theta in Quadrant IV
sin theta=
a. 3/5
b. 4/5
c. -4/5
d. -3/5

Answer 1

theta=pi n ? http://www.wolframalpha.com/input/?i=sec%5E2%28theta%29%3D1-tan%5E2%28theta%29

Answer 2

For this you are using wolfram alpha ???

Answer 3

Sorry:

\[\large \sec^2(\theta) = 1 + \tan^2 (\theta)\]

Answer 4

Oh I see. True. theta= 2 pi n + pi

Answer 5

\[\large \sec^2(\theta) = 1 + \tan^2 (\theta) = 1 + (\frac{-3}{4})^2 \implies \frac{25}{16}\]

\[\sec(\theta) = \frac{5}{4}\]

Answer 6

Cos will be reverse of this:

\[\cos(\theta )= \frac{4}{5}\]

Answer 7

Now use:

\[\sin^2 (\theta ) = 1 - \cos^2(\theta) = 1 - \frac{16}{25} = \frac{3}{5}\]

But your answer will be negative because \(sin(\theta)\) in 4rt quadrant is negative..

Answer 8

So:

\[\large \color{green}{\sin(\theta) = \frac{-3}{5}}\]

Answer 9

Oh, O.K. Thank you so much!

Answer 10

How would I find cot theta and csc theta for this problem?

Answer 11

@waterineyes

Answer 12

\[\cot(\theta)= \frac{1}{\tan(\theta)}\]

\[cosec(\theta) = \frac{1}{\sin(\theta)}\]

Answer 13

-4/3 and -5/3, can you please clarify? @waterineyes I really appreciate your help

Answer 14

Yes you are absolutely right..