Verify the identity:

(1 + tan^2x)(1 - sin^2x) = 1


I know the solution, but not the steps to get me the answers I need. I will need to know how to do this on my own for a test soon, so any explanations will be appreciated.


Solution, kinda: (1 + tan^2x)(1 - sin^2x) = sex^2x * cos^2x = 1/cos^2x = 1


Here are some identities that I'm pretty sure need to be used, but I'm not sure how or in what context to use them.


Quotient Identities:

tan x = (sin x)/(cos x)

cot x = (cos x)/(sin x)


Reciprocal Identities:

csc x = 1/(sin x)

sec x = 1/(cos x)

cot x = 1/(tan x)


Pythagorean Identities:

sin^2x + cos^2x = 1

sec^2x = 1 + tan^2x

csc^2x = 1 + cot^2x


Any and all help will be greatly appreciated!

Verified Answer :

\((1+\tan^2x)(1-\sin^2x)=\sec^2x(1-\sin^2x)\)

which follows from the Pythagorean identity. The same identity tells you that

\(\sec^2x(1-\sin^2x)=\sec^2x\cos^2x\)

Finally, since \(\sec x=\dfrac1{\cos x}\) (by definition), you're left with

\(\sec^2x\cos^2x=\dfrac1{\cos^2x}\times\cos^2x=1\)

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