Question

Verify the trigonometric equation by substituting identities to match the right hand side of the equation to the left hand side of the equation.

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

Answer 1

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Answer 2

please help

Answer 3

\(-tan^2x+sec^2x=1\)
\(\Longleftrightarrow ~sec^2x=1+tan^2x\)
\(\Longleftrightarrow ~ \Large \frac {1}{cos^2x} \normalsize= 1+ \Large \frac{sin^2s}{cos^2x}\)
\(\Longleftrightarrow ~ \Large \frac {1}{cos^2x} \normalsize=\Large \frac{cos^2x+sin^2s}{cos^2x}\) (true)

Answer 4

well use
- tan^2 = - sin^2/cos^2 and sec^2 = 1/cos^2

Answer 5

you have the common denominator... of cos^2

so you have (-sin^2 + 1)/cos^2

Answer 6

rewriting the numerator and you get \[\frac{1 - \sin^2(x)}{\cos^2(x)}\] use a well know trig identity and substitute it into the numerator...

Answer 7

I understand everything but the last part. What do you mean a well known trig identity?

Answer 8

well the best known trig identity is

\[\sin^2(x) + \cos^2(x) = 1\] just make cos^2 the subject and it should be obvious

Answer 9

I'm stuck at
(-sin^2x + 1)/cos^2x = 1

Answer 10

dang \[\sin^2(x) + \cos^2(x) = 1 ~~~~~then~~~~~\cos^2(x) = -\sin(x) + 1\]

Answer 11

substitute this into the numerator...

Answer 12

you mean denominator?

Answer 13

ohh

Answer 14

either substitution works... I'd substitute the numerator

Answer 15

I got it. Thank you so much.

Answer 16

and for my 3rd try its\[\sin^2(x) + \cos^2(x) = 1~~~then~~~~\cos^2(x) = -\sin^2(x) + 1\]