Question

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

Please explain. Thank you!

Answer 1

\[(a+b)^2 = a^2 + b^2 + 2ab\]

Answer 2

Use this formula..

Answer 3

Im still a little confused do I plug something into that formula

Answer 4

then you will need these three :
\[1. \tan(x) \cdot \cot(x) = 1 \\ 2. \tan^2(x) = \sec^2(x)-1 \\ 3. \cot^2(x) = cosec^2(x)-1\]

Answer 5

yes, put \(a = \tan(x)\) and \(b=\cot(x)\)..

Answer 6

so tan and cot = 1 so I plug one into the formula like this (tan (1) + cot (1))^2?

Answer 7

\[(\tan(x) + \cot(x))^2 = \tan^2(x) + \cot^2(x) + 2 \tan(x) \cdot \cot(x) \\ \implies \sec^2(x)-1 + cosec^2(x) - 1 + 2(1) \\ \implies \sec^2(x) + cosec^2(x) -\cancel{2} + \cancel{2} \\ \implies \sec^2(x) + cosec^2(x)\]

Answer 8

@waterineyes So the answer is B. sec^2 x + csc^2 x?

Answer 9

Ask yourself, not me.. :P