In this problem, we're asked to verify the identity by working with both sides of the equation. And the identity that we're given here is the secant of theta times one minus the sine squared of theta is equal to 1 plus the tangent squared of theta times the cotangent squared of theta divided by the cosecant squared of theta times the secant of theta. Now remember that whenever we're working on verifying an identity, we want to start with our more complicated side first, which is very clearly this right side because there is a lot going on there. So let's go ahead and get started and figure out how to simplify this. Now looking at this right side, 1 plus the tangent squared times the cotangent squared over the cosecant squared times the secant squared. Where can we go from here? Well remember we want to be constantly scanning for identities and the first thing that I notice here in my numerator is 1 plus the tangent squared. Now 1 plus the tangent squared, taking a look at my Pythagorean identities, is equal to the secant squared. So I can go ahead and replace that in my numerator there. I have the secant squared of theta times the cotangent squared of theta with that same denominator, the cosecant squared of theta times the secant of theta. Now here I can go ahead and just cancel this secant on the bottom with one of the secants at the top. So now I am left with the secant of theta times the cotangent squared of theta over the cosecant squared of theta. Now where can we go from here? Well remember, let's take a look at our strategies here. And here we can go ahead and break down everything in terms of sine and cosine because we can't add any fractions, and we don't have 1 plus or minus a trig function and we can't factor here either. So let's go ahead and break this down. Now I know that the secant of theta is equal to 1 over the cosine of theta and the cotangent squared of theta is equal to the cosine squared of theta over the sine squared of theta. Then in my denominator, I can also rewrite that in terms of sine. It is one over the sine squared of theta. So let's go ahead and work with this and see what we can cancel. now in that numerator, one of those cosines is going to cancel, and I'm just left with the cosine over the sine squared. And then looking at this, there's a lot of fractions going on. So let's go ahead and rewrite this so that you can clearly see what I'm gonna do here. So here I have the cosine of theta over the sine squared of theta. Now whenever we divide by a fraction, that's the same thing as multiplying by the reciprocal. So really, I'm multiplying this by the sine squared of theta over 1. Now here I have sine squared on the top, sine squared on the bottom, that goes away And all I'm left with is the cosine of theta. Now this looks really simplified. So let's go ahead and work with that left side now to see if we can get that to equal the cosine of theta. Now looking at my left side, the secant of theta times 1 minus the sine squared of theta, let's see what we can do here. We are constantly scanning for identities and looking here I have 1 minus the sine squared of theta, which I know is one of my Pythagorean identities just in a slightly different form because the sine squared of theta plus the cosine squared of theta is equal to 1. So if I go ahead and subtract the sine squared of theta from both sides, I am left with exactly what's in those parentheses there, and that will be equal to the cosine squared of theta. So I can rewrite this as the secant of theta times cosine squared of theta. Now where can we go from here? Well, the secant of theta I know is 1 over the cosine of theta. And if this is multiplying the cosine squared of theta, that's gonna be really helpful. So what can we do here? Well, I have a cosine on the bottom and a cosine on the top. And since that cosine was squared, I am still left with one of them, and all I'm left with here is the cosine of theta. Now these sides had dramatically different amounts of work that we had to do, but they're still equal to each other. So I can bring this all the way down here because we can clearly see that now this identity is verified. The cosine of theta is equal to the cosine of theta. Thanks for watching and I'll see you in the next one.
Table of contents
- 0. Review of College Algebra4h 43m
- 1. Measuring Angles39m
- 2. Trigonometric Functions on Right Triangles2h 5m
- 3. Unit Circle1h 19m
- 4. Graphing Trigonometric Functions1h 19m
- 5. Inverse Trigonometric Functions and Basic Trigonometric Equations1h 41m
- 6. Trigonometric Identities and More Equations2h 34m
- 7. Non-Right Triangles1h 38m
- 8. Vectors2h 25m
- 9. Polar Equations2h 5m
- 10. Parametric Equations1h 6m
- 11. Graphing Complex Numbers1h 7m
6. Trigonometric Identities and More Equations
Introduction to Trigonometric Identities
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