Question

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

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

Answer 1

Do you know this very famous identity?

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

Answer 2

Your problem is related to it.

Answer 3

binomial theorem?

Answer 4

Start with your problem.
Express tan and sec in terms of sin and cos.
Then multiply both sides by \(\cos^2 x\)

Answer 5

I'm not so sure on how to do that though @mathstudent55

Answer 6

Let's start with the tangent function.
We want to express the tangent as a function of sine and cosine.

Answer 7

From the trig of right triangles, you should have learned theses definitions of sine, cosine, and tangent.

\(\sin x = \dfrac{opp}{hyp} \)

\(\cos x = \dfrac{adj}{hyp}\)

\(\tan x = \dfrac{opp}{adj} \)

Also, the secant is 1/cosine, so

\(\sec x = \dfrac{hyp}{adj} \)

Answer 8

Now let's see what happens when you divide the sine by the cosine:

\(\large \dfrac{\sin x}{\cos x} = \dfrac{\frac{opp}{hyp}}{\frac{adj}{hyp}} = \dfrac{opp}{adj}\)

We know from the definition of tangent that \(\tan x = \dfrac{opp}{adj} \)

That means

\(\tan x = \dfrac{\sin x}{\cos x}\)

This is an identity that we need.

Answer 9

I know those I just don't know how to apply it here being that I'm not given a triangle with measurements.

Answer 10

Using the same type of manipulation of the definitions, you will get this other identity:

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

Answer 11

Ok, now we use these two identities for tan and sec in your problem. Where we see tan and sec, we replace them with the identities that give us tan and sec in terms of sine and cosine.

Answer 12

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

\(-\dfrac{\sin^2 x}{\cos^2 x} + \dfrac{1}{\cos^2} = 1\)

Ok so far?

Answer 13

Yes

Answer 14

Now multiply the entire equation by \(\cos^2 x\)

Answer 15

the 1 as well?

Answer 16

\(\cos^2 x\left(-\dfrac{\sin^2 x}{\cos^2 x} + \dfrac{1}{\cos^2 x} \right)= 1 \times \cos^2 x\)

We distribute \(\cos^2x \) on the left side and simplify each fraction.

\(-\dfrac{\sin^2 x \cos^2 x}{\cos^2 x} + \dfrac{\cos^2 x}{\cos^2 x} = \cos^2 x\)

\(-\dfrac{\sin^2 x \cancel{\cos^2 x}}{\cancel{\cos^2 x}~~1} + \dfrac{\cancel{\cos^2 x}~~1}{\cancel{\cos^2 x}~~1} = \cos^2 x\)

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

Now we add \(\sin^2 x \) to both sides:

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

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

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

This is the identity that I mentioned above that is known to be true.

Answer 17

I probably would have been a little lazier and just used the Pythagorean Identity that relates secant and tangent.\[\large\rm \color{orangered}{1+\tan^2x=\sec^2x}\]And then applying that to our problem:\[\Large\rm -\tan^2x+\color{orangered}{\sec^2x}=1\]\[\Large\rm -\tan^2x+\color{orangered}{1+\tan^2x}=1\]It cleans up pretty nicely, yes? :o

Answer 18

Thankyou!!!