Question

1/cot^2theta+sec theta cos theta

Answer 1

what do you need?

Answer 2

help

Answer 3

\[\frac{ 1 }{ \cot ^2 \theta }+\sec \theta \cos \theta\] is this what you have?

Answer 4

need to find expression as a constant a single trigo function

Answer 5

yea

Answer 6

so \[\frac{ 1 }{\cot \theta }=\tan \theta\]
and \[\sec \theta = \frac{ 1 }{ \cos \theta }\]
what does that get you?

Answer 7

how did u got this stuff

Answer 8

that's just the definition

Answer 9

i really dont get this question at all can u help me plz

Answer 10

\[\tan \theta = \frac{ \text{opposite} }{ \text{adjacent} }\text{, }\cot \theta = \frac{ \text{adjacent} }{ \text{opposite} }\Rightarrow \frac{ 1 }{ \cot \theta } = \tan \theta\]\[\sec \theta = \frac{ \text{hypoteneuse} }{ \text{opposite}}\text{, }\cos \theta = \frac{ \text{opposite} }{ \text{hypoteneuse}}\Rightarrow \sec \theta = \frac{1}{\cos \theta}\]

Answer 11

so substitute...
\[\frac{ 1 }{ \cot ^2 \theta }= \tan^2 \theta\]
and \[\sec \theta \cos \theta = \frac{ 1 }{ \cos \theta }\cos \theta = 1\]

then you have the identity that \[\tan^2\theta +1 = \sec^2 \theta\]

Answer 12

so thats the answer

Answer 13

hey u their

Answer 14

their?

Answer 15

tan2 theta +1=sec^2 theta is the answer right

Answer 16

i know

Answer 17

thanks buddy

Answer 18

you're welcome