Question

show sin^2x(1+sin^2x)=sec^2x-cos^2x

Answer 1

try substituting sin^2x = 1 - cos^2x in left hand side

Answer 2

then I get:\[1+\sec ^{2}\theta -\cos ^{2}\theta-\sec ^{2}\theta \cos ^{2}\theta \]

Answer 3

right - now replace cos^2x by 1 /sec^2x in the last term and what do u get?

Answer 4

OMG! Thankss!

Answer 5

lol - welcome

Answer 6

But when I opened up the brackets at the start it didn't work out, can you check why?

Answer 7

lol - to be perfectly honest i went wrong myself
thats why i abanded it - i'll have another look
how did u get your result?

Answer 8

I just followed your way, 1st sub. sin^2x = 1-cos^2x and then in the last term subing cos^2x by 1/sec^2x and then cancelling it outt

Answer 9

it can't be a good day for me because i can't get

1+sec^2θ−cos^2θ−sec^2θcos^2θ

Answer 10

i'll have to come back to u on this...

Answer 11

@sim123 try plugging in x = 30 degrees into each side of the identity and see what you get. i'll try it as well.

Answer 12

sin^2 30 (1 + sin^2 30) = 0.5^2(1 + 0.5^2) = 0.25 * 1.25 = 0.3125

Answer 13

sec^2 30 - cos^2 30 = 1.33333 - 0.75 = 0.583333

Answer 14

we've been hitting our heads against a brick wall - they are not equal