Is #tan^2x -= sec^2x - 1# an identity?

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Answer 1 · 2018-01-09T17:10:50

True

Start with the well known pythagorean identity:

# sin^2x + cos^2x -= 1 #

This is readily derived directly from the definition of the basic trigonometric functions #sin# and #cos# and Pythagoras's Theorem.

Divide both side by #cos^2x# and we get:

# sin^2x/cos^2x + cos^2x/cos^2x -= 1/cos^2x #

# :. tan^2x + 1 -= sec^2x #

# :. tan^2x -= sec^2x - 1#

Confirming that the result is an identity.

Answer 2 · 2018-01-09T17:12:45

Yes, #sec^2-1=tan^2x# is an identity.

We start from #sin^2x+cos^2x=1#.

Then divide everything by #cos^2x#

#sin^2x/cos^2x+cos^2x/cos^2x=tan^2x+1=1/cos^2x=sec^2x#

Rearrange to find #tan^2x#

#tan^2x=sec^2x-1#