Question

@AZ

Answer 1

healp._.

Answer 2

aha, I finally figured it out

Answer 3

okay, so re-write tan^2 (x) in terms of sin and cos

what do you get?

Answer 4

sin^2 x / cos^2x

Answer 5

good, now remember the identity

sin^2 x + cos^2 x = 1

well let's re-write sin^2 (x) as 1 - cos^2 (x)

so look at your entire LHS and distribute the other sin^2 x into that

Answer 6

so just looking at the LHS

\(\tan^2(x) \sin^2(x)\)

\(\dfrac{\sin^2(x)}{\cos^2(x)} \sin^2(x)\)

\(\dfrac{1- \cos^2(x)}{\cos^2(x)} \sin^2(x)\)

distribute the sin^2 x

Answer 7

i don't follow

Answer 8

you understand how we got the 1- cos^2 (x)

right?

Answer 9

identity thingy, ye

Answer 10

okay so before we did that we had in the numerator

sin^2(x) * sin^2(x)

but we changed one into (1-cos^2(x))

(1-cos^2(x)) * sin^2(x)
so now we want to multiply it out and so distribute the sin^2(x) into the parenthesis

Answer 11

sin^2 x - (sin^2 x cos^2 x)

Answer 12

`distribute the sin^2(x) into the parenthesis`
?

Answer 13

good good

now we have

\(\dfrac{\sin^2( x )- \sin^2 (x) \cos^2 (x)}{\cos^2(x)} = \tan^2(x) - \sin^2(x)\)

and just remember the fraction rules
\(\dfrac{a-b}{c} = \dfrac{a}{c} -\dfrac{b}{c}\)

Answer 14

\(\color{#0cbb34}{\text{Originally Posted by}}\) @snowflake0531
`distribute the sin^2(x) into the parenthesis`
?
\(\color{#0cbb34}{\text{End of Quote}}\)

that's what you just did

(1 - cos^2x) * sin^2x

= sin^2 x - (sin^2 x cos^2 x)

Answer 15

OHHHHHHHHHHHHHHHH got it
THANKSSSSSSSSSSSSSSS <3

Answer 16

YOU'RE WELCOME :D

Answer 17

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