Question

Any body need help on cacle finding sec an csc

Answer 1

Sure bro.

Answer 2

Let me type out a nice little lesson.

Answer 3

\[\sec(\theta) = \frac{1}{\cos(\theta)} \ \ \ \ \ \ \ \ \ \csc(\theta) = \frac{1}{\sin(\theta)}\]\(\cos\) is the ratio of adjacent to hypotenuse.
\(\sin\) is the ratio of opposite to hypotenuse.

Answer 4

Okay, so what if you are given \(\sin(\theta) = \dfrac{3}{5}\)? It is clear that we have this triangle:

Now, use the Pythagorean Theorem. You have \(4\) as the remaining side.

Answer 5

So, \(\cos(\theta) = \dfrac{4}{5}\). Using the formulae initially given to you, you get:\[\boxed{\sec(\theta)= \dfrac{5}{4}} \rm \ \ and \ \ \boxed{\csc(\theta) = \dfrac{5}{3}} \]

Answer 6

Use the example to do your problem.

Answer 7

reciprocals all you need to know.