Use the Pythagorean identities to rewrite the expression with no fraction.11−secθ\frac{1}{1-\sec\theta}1−secθ1

− cot ⁡ 2 θ ( 1 + sec ⁡ θ ) -\cot^2\theta\left(1+\sec\theta\right) − cot 2 θ ( 1 + sec θ )

Use the Pythagorean identities to rewrite the expression with no fraction. 1 1 − sec θ \frac{1}{1-\sec\theta} 1 − s e c θ 1

In this problem, we're asked to use the Pythagorean identities in order to rewrite the expression with no fraction. Now the expression that we're given here is 1 over 1 minus the secant of theta. So how can we use our pythagorean identities here? Well, remember, if I have something like 1 minus some trig function, if I multiply that times 1 plus that same trig function, I'll end up with a pythagorean identity. So here, I wanna multiply this 1 minus the secant of theta times 1 plus the secant of theta. Now since I have a fraction here I want to make sure that I'm multiplying the top and the bottom by the same thing so as not to change the value of the expression that I'm working with. I'm really just multiplying it by 1. Now multiplying this out in my numerator I'm left with 1 plus the secant of theta. Then in my denominator multiplying this recognizing that it's a difference of squares, I am left with 1 minus the secant squared of theta. Now I have a squared trig function. And looking at my pythagorean identities, I want to use this second pythagorean identity in order to get this simplified. So here I want to replace this one minus the secant squared of theta, but it's not quite in the right form yet. So if I subtract the secant squared of theta from both sides, it will cancel on that right side. And then if I also subtract the tangent squared of theta from both sides, doing both of these steps at the same time, that will cancel on that left side. Now all I'm left with is 1 minus the secant squared of theta and that's equal to the negative tangent squared of theta. Now I can replace my denominator here using that pythagorean identity. So here in my numerator, I still have 1 plus the secant of theta but in my denominator, I have negative tangent squared of theta. Now we still have a fraction. So how do we get rid of that fraction? Well, remember that one over the tangent of theta is simply equal to the cotangent of theta. So one over the tangent squared of theta is going to be the cotangent squared of theta. Now we don't want to forget this negative here. So we have negative cotangent squared of theta and that is multiplying what's actually in that numerator, 1 plus the secant of theta. Now that's a bit complicated but remember to take your time with each of these steps and work through these as much as you need to and practice as much as you can in order to get this down. Our final answer here with no fraction and our answer is negative cotangent squared of theta times 1 plus the secant of theta. Let me know if you have any questions and thanks for watching.

Use the Pythagorean identities to rewrite the expression with no fraction. 1 1 − sec ⁡ θ \frac{1}{1-\sec\theta} 1 − s e c θ 1

− cot ⁡ 2 θ ( 1 + sec ⁡ θ ) -\cot^2\theta\left(1+\sec\theta\right) − cot 2 θ ( 1 + sec θ )

1 + sec ⁡ θ 1+\sec\theta 1 + sec θ

1 tan ⁡ 2 θ \frac{1}{\tan^2\theta} t a n 2 θ 1

− tan ⁡ 2 θ ( 1 + sec ⁡ θ ) -\tan^2\theta\left(1+\sec\theta\right) − tan 2 θ ( 1 + sec θ )

12. Trigonometric Identities

Introduction to Trigonometric Identities

Precalculus

12. Trigonometric Identities

Introduction to Trigonometric Identities

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Multiple Choice

Use the Pythagorean identities to rewrite the expression with no fraction.
$\frac{1}{1-\sec\theta}$

$1+\sec\theta$

$\frac{1}{\tan^2\theta}$

$-\cot^2\theta\left(1+\sec\theta\right)$

$-\tan^2\theta\left(1+\sec\theta\right)$

Verified step by step guidance

Start by recognizing the Pythagorean identity: $ \sec^2\theta = 1 + \tan^2\theta $. This identity will help us rewrite expressions involving $ \sec\theta $ and $ \tan\theta $.

Rewrite $ \frac{1}{1 - \sec\theta} $ using the identity $ \sec\theta = \frac{1}{\cos\theta} $. This gives $ \frac{1}{1 - \frac{1}{\cos\theta}} = \frac{\cos\theta}{\cos\theta - 1} $.

Next, consider $ \frac{1}{\tan^2\theta} $. Using the identity $ \tan^2\theta = \sec^2\theta - 1 $, rewrite $ \frac{1}{\tan^2\theta} $ as $ \frac{1}{\sec^2\theta - 1} $.

Combine the expressions: $ \frac{\cos\theta}{\cos\theta - 1} \cdot \frac{1}{\sec^2\theta - 1} \cdot (1 + \sec\theta) $. Simplify using the identity $ \sec^2\theta = 1 + \tan^2\theta $.

Finally, simplify the expression to get $ -\cot^2\theta(1 + \sec\theta) $ or $ -\tan^2\theta(1 + \sec\theta) $ by recognizing the relationship between $ \tan\theta $ and $ \cot\theta $.

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