Simplify the expression.tan(−θ)sec(−θ)\frac{\tan\left(-\theta\right)}{\sec\left(-\theta\right)}sec(−θ)tan(−θ)

− sin ⁡ θ -\sin\theta − sin θ

Simplify the expression. tan ( − θ ) sec ( − θ ) \frac{\tan\left(-\theta\right)}{\sec\left(-\theta\right)} s e c ( − θ ) t a n ( − θ )

Here we're asked to simplify the expression, the tangent of negative theta over the secant of negative theta. Now remember that in order for a true expression to be fully simplified, all of our arguments must be positive. And currently, both of our arguments are negative. So that's the first thing that we wanna fix. Now remembering that we have our even odd identities down here for whenever we have a negative argument. So here I see that the tangent of negative theta is equal to the negative tangent of theta, And the secant of negative theta is just going to be equal to the secant of theta because it is an even function just like cosine. So let's go ahead and rewrite this using those identities to give these positive arguments. So this is simply negative tangent of theta over the secant of theta. Now we have positive arguments here, but we don't want any fractions, and we want as few trig functions as possible. So what else can we do here? Well, remember one of our strategies is going to be to break down everything in terms of sine and cosine. And here I have the tangent and the secant, both of which can be rewritten using identities to get them in terms of sine and cosine. So the tangent, negative tangent here, becomes negative sine of theta over the cosine of theta. Then in my denominator, the secant of theta is just one over the cosine of theta. Now this looks a little bit scary because there's a fraction on top of another fraction. But remember that whenever we're dividing by a fraction, we're really just multiplying by the reciprocal. So this is really just the negative sine of theta over the cosine of theta times the cosine of theta over 1, flipping that bottom fraction. Now we can cancel some stuff because I have a cosine on the top and a cosine on the bottom, leaving me with just negative sine of theta divided by 1, which is just negative sine of theta. So now I have no fractions and I have as few trig functions as possible because I just have one. So we have fully simplified this expression down to the negative sine of theta. Let me know if you have any questions. Thanks for watching.

Simplify the expression. tan ⁡ ( − θ ) sec ⁡ ( − θ ) \frac{\tan\left(-\theta\right)}{\sec\left(-\theta\right)} s e c ( − θ ) t a n ( − θ )

− sin ⁡ θ -\sin\theta − sin θ

− cot ⁡ θ -\cot\theta − cot θ

6. Trigonometric Identities and More Equations

Introduction to Trigonometric Identities

Trigonometry

6. Trigonometric Identities and More Equations

Introduction to Trigonometric Identities

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Multiple Choice

Simplify the expression.
$\frac{\tan\left(-\theta\right)}{\sec\left(-\theta\right)}$

$\sin\theta$

$-\sin\theta$

$-\cot\theta$

Verified step by step guidance

Recognize that $ \tan(-\theta) = -\tan(\theta) $ and $ \sec(-\theta) = \sec(\theta) $ due to the even-odd properties of trigonometric functions.

Substitute these identities into the expression: $ \frac{\tan(-\theta)}{\sec(-\theta)} \sec(-\theta) \tan(-\theta) $. This becomes $ \frac{-\tan(\theta)}{\sec(\theta)} \sec(\theta) (-\tan(\theta)) $.

Simplify the expression by canceling out $ \sec(\theta) $ in the numerator and denominator: $ -\tan(\theta) \cdot (-\tan(\theta)) $.

Recognize that $ \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} $, so $ -\tan(\theta) \cdot (-\tan(\theta)) = \left(\frac{\sin(\theta)}{\cos(\theta)}\right)^2 $.

Simplify further: $ \left(\frac{\sin(\theta)}{\cos(\theta)}\right)^2 = \frac{\sin^2(\theta)}{\cos^2(\theta)} = \tan^2(\theta) $.

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