Question

The expression (tan x + cot x)^2 is the same as ____?

Please help and explain. Thank you!

Answer 1

(a+b)^2=a^2+2ab+b^2

Answer 2

What do I do with this formula do I plug something in?

Answer 3

you have (a+b)^2 where a=tan(x) and b=cot(x) don't you?

Answer 4

\[(\tan(x)+\cot(x))^2\\ (\tan(x)+\cot(x)) \cdot (\tan(x)+\cot(x)) \\ \tan(x)(\tan(x)+\cot(x))+\cot(x)(\tan(x)+\cot(x)) \\ \tan^2(x)+\tan(x)\cot(x)+\cot(x)\tan(x)+\cot^2(x)\]
are you can take this longer route if you don't know (a+b)^2=a^2+2ab+b^2

Answer 5

combine like terms there

Answer 6

\[\huge\ (a+b)^2 = (a + b)(a+b)\]
this is example now use foil method

Answer 7

also you should know a little fact about tan(x)*cot(x)

Answer 8

you might also find the Pythagorean identities helpful here too

Answer 9

\[\sin^2(x)+\cos^2(x)=1 \\ \tan^2(x)+1=\sec^2(x) \\ 1+\cot^2(x)=\csc^2(x)\]

Answer 10

@freckles The answer looks like it is B. sec^2 x + csc^2 x

Answer 11

sounds right
tan^2(x)+1+1+cot^2(x)

sec^2(x) + csc^2(x)

:)