Question

Decide whether or not cos^2 x(1+tan^2 x) = 1. Explain your reasoning

Answer 1

\[Cos^2(x) * (1+\tan^2(x))\]

Is this your question?

Answer 2

\(cos^2 x(1+tan^2 x) = 1\)

Let's use this identity (in red, below):

\(\color {red}{\sin^2 x + cos^2 x = 1}\)

Divide both sides by \( \cos^2 x\).

\(\color{red}{\tan^2 x + 1 = \sec^2 x}\)

Solve for \(\tan^2 x\).

\(\color{red}{\tan^2 x = \sec^2 x - 1}\)

Now substitute what \(\tan^2 x \) is equal to in the original equation.

\(cos^2 x(1+\color{red}{\sec^2 x - 1}) = 1\)

Can you continue from here?

Answer 3

I'm still not sure. What am I looking for to see if it is an identity or not?

Answer 4

To prove that an equation is an identity, you must show that the left side and right side equal each other.

Answer 5

so I have to make the left side = 1?

Answer 6

Correct.
We're almost there.
Subtract the 1 and 1 in the parentheses.
What do you have now?

Answer 7

cos^2 x(sec^2 x)=1

Answer 8

What is \( \sec^2 x\) equal to? (in terms of \( \cos^2 x\))

Answer 9

hmm not sure. tan^2x

Answer 10

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

Answer 11

ok, so it is a trig identity right?

Answer 12

Correct, because the left side equals 1.

Answer 13

Ah ok so to clarify: I would just explain the reasoning by writing both sides have to be equal and show that sec^2 x = 1/cos^2x

Answer 14

I just put the equation into symbolab and it said the identity was false? @mathstudent55

Answer 15

@mathstudent55

Answer 16

Who is "it"?

Answer 17

http://www.wolframalpha.com/input/?i=sec^2+x+%3D+1%2Fcos^2+x

Answer 18

http://www.wolframalpha.com/input/?i=cos^2+x%281%2Btan^2+x%29+%3D+1

Answer 19

Hmm that's weird then! I was just confused because the program said it was false. Thanks for explaining and linking me to Wolf

Answer 20

1. Given:
\[\cos^2 x(1+\tan^2 x) = 1\]

2. Re-write \(\tan^2x(x)\) as \(\dfrac{\sin^2x}{\cos^2x}\):

\[\cos^2x\left(1 + \frac{\sin^2x}{\cos^2x}\right) = 1\]

3. Factor out \(\dfrac{1}{\cos^2x}\):
\[\frac{\cos^2x}{\cos^2x} \left( \cos^2x + \sin^2x \right) = 1\]

4. From there, it is easy to see that:
\[1(1) = 1\]