Question

find the value of cot^2(2tan^(-1)(4/3))

Answer 1

\[\cot^2(2 \tan^{-1}(\frac{4}{3}))\]
\[\text{ Let } y=\tan^{-1}((\frac{4}{3}))\]

=>
\[\tan(y)=\frac{4}{3}\]

The hyp. is 5.
\[\cot^2(2 \cdot y)=[\cot(2y)]^2=[\frac{\cos(2y)}{\sin(2y)}]^2=[\frac{\cos^2(y)-\sin^2(y)}{2 \sin(y) \cos(y)}]^2\]
\[=[\frac{[\cos(y)]^2-[\sin(y)]^2}{2 \sin(y) \cos(y)}]^2\]
\[=[\frac{[\frac{3}{5}]^2-[\frac{4}{5}]^2}{2 \cdot \frac{4}{5} \cdot \frac{3}{5}}]^2\]
\[=[\frac{\frac{9}{25}-\frac{16}{25}}{2 \cdot \frac{12}{25}}]^2\]
\[=[\frac{9-16}{2 \cdot 12}]^2\]
\[=[\frac{-7}{24}]^2=\frac{49}{24^2}\]

Answer 2

can you help me with the rest of the problem?

Answer 3

What do you mean? Doing 24^2?

Answer 4

24^2 means 24*24
You can do the multiplication. I believe in you.

Answer 5

hahahaha no i know how to do that but the whole problem was
sec(tan^-1(-4/3)-csc^-1(-13/12))cot^2(2tan^-1(4/3)

Answer 6

It is hard for me to read that
I did \[cot^2(2 \cdot \tan^{-1}(\frac{4}{3}))\]
Which is the problem I read above? Is that not the right problem?

Answer 7

yeah you did the last part of the problem the part you have to multiply with the first part

Answer 8

\[\sec(\tan^-1(-4/3)-\csc^-1(-13/12))\cot^2(2\tan^-1(4/3))\]
is that better?

Answer 9

\[\sec(\tan^{-1}(\frac{-4}{3})-\csc^{-1}(\frac{-13}{12})) ?\]
So you are saying this is the first part?

Answer 10

yes and what you solved was the second part

Answer 11

Which was the only part you had lol
But ok...

Answer 12

So we have
Let y=arctan(-4/3) and let x=arccsc(-13/12)
Then we have
\[\sec(y-x)=\frac{1}{\cos(y-x)}=\frac{1}{\cos(y)\cos(x)+\sin(y)\sin(x)}\]

Answer 13

So draw two right triangles for both of those Let parts...