MHB Angle Sum/Difference Identities: Billy's Pre-calc Math Problem

[MarkFL]

Here is the question:

Pre-calc math problem?

a and B are quadrent I angles with cos(a) = 15/17 and csc(B) = 41/9.

find tan (a + B) and tan (a-B)

Here is a link to the question:

Pre-calc math problem? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

 

[MarkFL]

Re: Billy's question from Yahoo! Answers involving the angle sum/difference identities for tangent

Hello Billy,

We are given:

$$\cos(\alpha)=\frac{15}{17}$$

and using the Pythagorean identity $\tan^2(\alpha)=\sec^2(\alpha)-1$ we find (given $\alpha$ is in the first quadrant, and so all trig. functions are positive there:

$$\tan(\alpha)=\sqrt{\left(\frac{17}{15} \right)^2-1}=\frac{8}{15}$$

We are also given:

$$\csc(\beta)=\frac{41}{9}$$

and using the Pythagorean identity $\cot^2(\beta)=\csc^2(\beta)-1$ we find:

$$\tan(\beta)=\frac{1}{\cot(\beta)}=\frac{1}{ \sqrt{\left(\frac{41}{9} \right)^2-1}}=\frac{9}{40}$$

Now, using the angle sum/difference identity for tangent $$\tan(\alpha\pm\beta)=\frac{\tan(\alpha)\pm\tan( \beta)}{1\mp\tan(\alpha)\tan( \beta)}$$, we find:

$$\tan(\alpha+\beta)=\frac{\frac{8}{15}+\frac{9}{40}}{1-\frac{8}{15}\cdot\frac{9}{40}}=\frac{455}{528}$$

$$\tan(\alpha-\beta)=\frac{\frac{8}{15}-\frac{9}{40}}{1+\frac{8}{15}\cdot\frac{9}{40}}= \frac{185}{672}$$

To Billy and any other visitors reading this topic, I invite you to register and post other trigonometry questions in our http://www.mathhelpboards.com/f12/ forum.