Question

pick 4 of the following identities to verify. show all work. write question number and show all work in the boxes provided. put a check on the final step showing verification.

1. \\(\frac{\csc\theta}{\sec\theta} + \frac{\cos\theta}{\sin\theta} = 2\cot\theta\\) 9. \\(1 - \frac{\sin^2\theta}{1 - \cos\theta} = -\cos\theta\\) 10. \\(\frac{\cot(-\theta)}{\csc\theta} = -\cos\theta\\) 11. \\(\sin\theta + \cos\theta\cot\theta = \csc\theta\\)
2. \\(\frac{\cos\theta}{1 - \sin\theta} = \sec\theta + \tan\theta\\) 13. \\(\frac{\sin\theta}{\csc\theta} + \frac{\cos\theta}{\sec\theta} = 1\\) 14. \\(\frac{1 + \cos\theta}{\sin\theta} + \frac{\sin\theta}{1 + \cos\theta} = 2\csc\theta\\) 15. \\(\cos\theta\sec^2\theta\tan\theta - \cos\theta\tan^3\theta = \sin\theta\\)

Explanation:

Let's verify identity #8: $\boldsymbol{\frac{\csc\theta}{\sec\theta} + \frac{\cos\theta}{\sin\theta} = 2\cot\theta}$

Step 1: Rewrite using reciprocal identities

Recall: $\csc\theta = \frac{1}{\sin\theta}$, $\sec\theta = \frac{1}{\cos\theta}$, and $\cot\theta = \frac{\cos\theta}{\sin\theta}$.
Substitute $\csc\theta$ and $\sec\theta$ in the first term:
$$\frac{\csc\theta}{\sec\theta} = \frac{\frac{1}{\sin\theta}}{\frac{1}{\cos\theta}} = \frac{\cos\theta}{\sin\theta}$$

Step 2: Simplify the left-hand side (LHS)

The second term is already $\frac{\cos\theta}{\sin\theta}$ (which is $\cot\theta$). So:
$$\text{LHS} = \frac{\cos\theta}{\sin\theta} + \frac{\cos\theta}{\sin\theta}$$

Step 3: Combine like terms

Add the two identical fractions:
$$\frac{\cos\theta}{\sin\theta} + \frac{\cos\theta}{\sin\theta} = 2 \cdot \frac{\cos\theta}{\sin\theta}$$

Step 4: Verify with $\cot\theta$

Since $\cot\theta = \frac{\cos\theta}{\sin\theta}$, substitute:
$$2 \cdot \frac{\cos\theta}{\sin\theta} = 2\cot\theta$$

Final Check:

LHS simplifies to $2\cot\theta$, which matches the right-hand side (RHS). $\boldsymbol{\checkmark}$

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Let's verify identity #11: $\boldsymbol{\sin\theta + \cos\theta \cot\theta = \csc\theta}$

Step 1: Rewrite $\cot\theta$ using $\frac{\cos\theta}{\sin\theta}$

Recall: $\cot\theta = \frac{\cos\theta}{\sin\theta}$. Substitute into the second term:
$$\cos\theta \cot\theta = \cos\theta \cdot \frac{\cos\theta}{\sin\theta} = \frac{\cos^2\theta}{\sin\theta}$$

Step 2: Simplify the left-hand side (LHS)

Rewrite $\sin\theta$ with a common denominator ($\sin\theta$):
$$\sin\theta = \frac{\sin^2\theta}{\sin\theta}$$
Now add the two terms:
$$\text{LHS} = \frac{\sin^2\theta}{\sin\theta} + \frac{\cos^2\theta}{\sin\theta}$$

Step 3: Combine fractions and use Pythagorean identity

Combine over the common denominator:
$$\text{LHS} = \frac{\sin^2\theta + \cos^2\theta}{\sin\theta}$$
By the Pythagorean identity, $\sin^2\theta + \cos^2\theta = 1$:
$$\text{LHS} = \frac{1}{\sin\theta}$$

Step 4: Verify with $\csc\theta$

Since $\csc\theta = \frac{1}{\sin\theta}$, we have:
$$\frac{1}{\sin\theta} = \csc\theta$$

Final Check:

LHS simplifies to $\csc\theta$, which matches the right-hand side (RHS). $\boldsymbol{\checkmark}$

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Let's verify identity #13: $\boldsymbol{\frac{\sin\theta}{\csc\theta} + \frac{\cos\theta}{\sec\theta} = 1}$

Step 1: Rewrite using reciprocal identities

Recall: $\csc\theta = \frac{1}{\sin\theta}$ (so $\frac{1}{\csc\theta} = \sin\theta$) and $\sec\theta = \frac{1}{\cos\theta}$ (so $\frac{1}{\sec\theta} = \cos\theta$).
Substitute into each term:
$$\frac{\sin\theta}{\csc\theta} = \sin\theta \cdot \sin\theta = \sin^2\theta$$
$$\frac{\cos\theta}{\sec\theta} = \cos\theta \cdot \cos\theta = \cos^2\theta$$

Step 2: Simplify the left-hand side (LHS)

Add the two terms:
$$\text{LHS} = \sin^2\theta + \cos^2\theta$$

Step 3: Verify with Pythagorean identity

By the Pythagorean identity, $\sin^2\theta + \cos^2\theta = 1$.

Final Check:

LHS simplifies to $1$, which matches the right-hand side (RHS). $\boldsymbol{\checkmark}$

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Let's verify identity #14: $\boldsymbol{\frac{1 + \cos\theta}{\sin\theta} + \frac{\sin\theta}{1 + \cos\theta} = 2\csc\theta}$

Answer:

Step 1: Find a common denominator

The common denominator for $\sin\theta$ and $1 + \cos\theta$ is $\sin\theta(1 + \cos\theta)$. Multiply each fraction by the missing factor:
$$\frac{(1 + \cos\theta)(1 + \cos\theta)}{\sin\theta(1 + \cos\theta)} + \frac{\sin\theta \cdot \sin\theta}{\sin\theta(1 + \cos\theta)}$$

Step 2: Expand and simplify the numerator

Expand the first numerator: $(1 + \cos\theta)^2 = 1 + 2\cos\theta + \cos^2\theta$.
The second numerator: $\sin^2\theta$.
Combine numerators over the common denominator:
$$\frac{1 + 2\cos\theta + \cos^2\theta + \sin^2\theta}{\sin\theta(1 + \cos\theta)}$$

Step 3: Use the Pythagorean identity

Recall $\sin^2\theta + \cos^2\theta = 1$. Substitute into the numerator:
$$1 + 2\cos\theta + (\cos^2\theta + \sin^2\theta) = 1 + 2\cos\theta + 1 = 2 + 2\cos\theta$$

Step 4: Factor and simplify

Factor the numerator: $2 + 2\cos\theta = 2(1 + \cos\theta)$.
Now the expression becomes:
$$\frac{2(1 + \cos\theta)}{\sin\theta(1 + \cos\theta)}$$

Step 5: Cancel common terms

Cancel $(1 + \cos\theta)$ from numerator and denominator:
$$\frac{2}{\sin\theta}$$

Step 6: Verify with $\csc\theta$

Since $\csc\theta = \frac{1}{\sin\theta}$, substitute:
$$\frac{2}{\sin\theta} = 2\csc\theta$$

Final Check:

LHS simplifies to $2\csc\theta$, which matches the right-hand side (RHS). $\boldsymbol{\checkmark}$

(You can choose any 4 identities—these are examples of verification!)