How would you solve these? Cot^-1(-sq. 3/3), cos(tan^-1 u)

[Violagirl]

1. Cot^-1(-sq. 3/3)

2. cos(tan^-1 u)

I kept getting lost trying to solve these:

1. cos theta/1-tan theta + sin theta/1 - cot theta = sin theta + cos theta

I was able to start by converting the denominators to make it:

cos theta/1-sin theta/cos theta + sin theta/1-cos theta/sin theta but am stuck after that.

2. tan theta+sec theta-1/tan theta-sec theta+1=tan theta + sec theta.

Thanks a ton!

 

[stapel]

1. Cot^-1(-sq. 3/3)

(I will use "@" to indicate the angle "theta".)

By definition, the above means that cot(@) = -sqrt[3]/3, so tan(@) = -3/sqrt[3] = -sqrt[3]. What then must the measure of the angle @?

2. cos(tan^-1 u)

Draw a right triangle; the particular scale doesn't matter. Note that "tan[sup:37j2bi15]-1[/sup:37j2bi15](u) = @" means that tan(@) = u = u/1. Label the "opposite" as "u" and the "adjacent" as "1". Use the Pythagorean Theorem to find an expression for the length of the hypotenuse. Then read off "adjacent over hypotenuse" to find the expression for the cosine.

1. cos theta/1-tan theta + sin theta/1 - cot theta = sin theta + cos theta

2. tan theta+sec theta-1/tan theta-sec theta+1=tan theta + sec theta.

Are you supposed to be "solving" these equations, or are they identities that you need to prove?

Thank you!

 

[Deleted member 4993]

1. Cot^-1(-sq. 3/3)

2. cos(tan^-1 u)

I kept getting lost trying to solve these:

1. cos theta/1-tan theta + sin theta/1 - cot theta = sin theta + cos theta

I was able to start by converting the denominators to make it:

cos theta/1-sin theta/cos theta + sin theta/1-cos theta/sin theta but am stuck after that.

2. tan theta+sec theta-1/tan theta-sec theta+1=tan theta + sec theta.

Thanks a ton!

2. tan theta+sec theta-1/tan theta-sec theta+1=tan theta + sec theta.

if your problem is:

Prove the identity:

\(\displaystyle \frac{\tan(\theta) \, +\, \sec(\theta) \, - \, 1}{\tan(\theta) \, -\, \sec(\theta) \, + \, 1} \, = \, \tan(\theta) \, +\, \sec(\theta)\)

you should have written it as:

Prove the identity:

(tan theta+sec theta-1)/(tan theta-sec theta+1) = tan theta + sec theta.

If it is then

\(\displaystyle \frac{\tan(\theta) \, +\, \sec(\theta) \, - \, 1}{\tan(\theta) \, -\, \sec(\theta) \, + \, 1}\)

\(\displaystyle = \, \frac{\tan(\theta) \, +\, [\sec(\theta) \, - \, 1]}{\tan(\theta) \, -\,[ \sec(\theta) \, - \, 1]}\)

multiply top and bottom of the LHS by (the conjugate of the denominator) :

\(\displaystyle \frac{\tan(\theta) \, +\, [\sec(\theta) \, - \, 1]}{\tan(\theta) \, +\, [\sec(\theta) \, - \, 1]}\)