Question

-tan^2x+sec^2x i know the answer is 1 but i need to prove it

Answer 1

@ganeshie8 a little help please

Answer 2

change everything into `sin` and `cos`

Answer 3

\[(-\sin^2/\cos^2x)+(1/\cos^2x)\]

Answer 4

right this is what i got

Answer 5

Yes! combine the fractions

Answer 6

(-sin^2+1)/cos^2x

Answer 7

and i just get stuck here

Answer 8

\[\large -\dfrac{\sin^2x}{\cos^2x} + \dfrac{1}{\cos^2x}\]

\[\large \dfrac{-\sin^2x+1}{\cos^2x}\]

Answer 9

next use below identity :
\[1 = \sin^2x + \cos^2x\]

Answer 10

right that's the identity i thought i would need to use but i don't know how to apply it though

Answer 11

\[\large -\dfrac{\sin^2x}{\cos^2x} + \dfrac{1}{\cos^2x}\]

\[\large \dfrac{-\sin^2x+\color{red}{1}}{\cos^2x}\]

\[\large \dfrac{-\sin^2x+\color{red}{\sin^2x + \cos^2x}}{\cos^2x}\]

Answer 12

ok i see now so -sin^2 and sin^2x cancels out, then cos^2x/cos^2x will be one

Answer 13

thats it!

Answer 14

thank you so much! :)

Answer 15

np:)