Question

Verify the identity. Please help :)

Answer 1

you can cross multiply, giving (1-sinx)(1+sinx) = cos^2x
then use the difference of squares identity
1-sin^2x=cos^2x
now
1=sin^2x+cos^2x
which is the fundamental identity of the two trig functions. QED

Answer 2

Thank you so much, i figured it was cross multiplying.. I really appreciate it :)

Answer 3

no prob

Answer 4

Cross multiplying is incorrect here!
be aware! invalid way of doing this
if we knew that they indeed equal we would not ask you to verify them in the first place
the valid way is to start from one side and get the other side

Answer 5

we can cross multiply and show that this reduces to an identity. This is just a rearranged identity, so we in fact can do it.

Answer 6

No, logically it is not correct! remember math is following certain logical ways to derive thing and achieve results
if you are doing computer this would not make any sense

Answer 7

that is a common mistake many people do to get the result
to prove something start always with one side
and achieve the other side
here is the issue with the way you did it
we want to prove A=B
you started with A=B then at the end you found that 1=1
this is not logically accepted you started with what we want to prove to get 1=1
what you should do is start from what we know 1=1 and go from there
not backwards

Answer 8

for your question:
start from the left side
multiply top and bottom by 1+sinx
\(\large \frac{1-\sin x}{\cos x}=\large \frac{(1-\sin x)(1+\sin x)}{\cos x(1+\sin x)}\)

Answer 9

it becomes \(\Large{ \frac{1-\sin^2 x}{\cos x(1+\sin x)}}=\Large \frac{\cos^2 x}{\cos x(1+\sin x)}\)

Answer 10

cancel cos x from top and bottom you get
\(\huge \frac{\cos x}{1+\sin x}\) Done!

Answer 11

@xapproachesinfinity oh my god, thank you so much.. I didn't see this until now!

Answer 12

yw!