Question

Verify the trig function:
can someone explain to me how my professor got from step 1 to 2? will give medals!

Answer 1

i attached a photo of the problem

Answer 2

@freckles @Nnesha

Answer 3

so he is working on the right side

Answer 4

yes

Answer 5

i'm sure you're familiar with the special identity \[\huge\rm \sin^2x+\cos^2x=1\]
solve this identity for sin^2x

Answer 6

1-cos^2x=sin^2x

Answer 7

right so now we can replace sin^2x with 1-cos^2
\[\huge\rm \tan^2x \color{reD}{sin^2x} \]
\[\huge\rm \tan^2x \color{reD}{(1-cos^2x)} \]

Answer 8

right

Answer 9

but how did they go from that ^ to tan^2-(sin^2/cos^2)?

Answer 10

that's what i'm trying to understand .... ,-,

Answer 11

can you take screenshot of next step

Answer 12

he said something about rewriting it as tan, but it does make sense

Answer 13

doesnt *

Answer 14

it doesn't make sense
i guess he is trying to rewrite tan^2 in terms of sin and cos

Answer 15

\[\huge\rm \tan^2x \color{reD}{(1-cos^2x)} \]\[\large\rm \frac{ \sin^2x }{ \cos^2x }(1-\cos^2x)\]

Answer 16

im still confused :(

Answer 17

hm let me think

Answer 18

ahh i see so he distributed by sin^2/cos^2
\[\rm \frac{ \sin^2 x}{ \cos^2x }-\frac{\sin^2}{\cos^2x}*\cos^2x\]
and then he changed sin^2/cos^2 to tan^2

Answer 19

im sorry, could you rewrite it out from the start, im still confused

Answer 20

\[\huge\rm \tan^2x \color{reD}{(1-cos^2x)} \]\[\large\rm \color{blue}{ \frac{ \sin^2x }{ \cos^2x} }(1-\cos^2x)\]
rewrite tan as sin/cos
\[\rm \color{blue}{ \frac{ \sin^2 x}{ \cos^2x }}-\frac{\sin^2}{\cos^2x}*\cos^2x\]
distribute parentheses by sin^2/cos^2
\[\rm \color{blue}{\tan^2x} -\frac{sin^2x}{cos^2x}*cos^2x\]
rewrite sin^2/cos^2 as tan^2

Answer 21

im still lost on this, im sorry

Answer 22

ohhh wait i think i get ittt

Answer 23

he distributed tan from the get go?

Answer 24

okay
so i applied distributive property \[\large\rm \color{ReD}{a}(b+c)=\color{ReD}{a}*b+\color{ReD}{a}*c=\color{red}{a}b+\color{ReD}{a}c\]
multiply both terms in the parentheses by outside term
that's how

\[\rm \color{blue}{ \frac{ \sin^2 x}{ \cos^2x }}-\frac{\sin^2}{\cos^2x}*\cos^2x\]