William W. answered • 12/12/20

Experienced Tutor and Retired Engineer

The second one is almost the same, just turn tan into sin/cos and cot into cos/sin and it works out the same.

Valerie G.

asked • 12/12/201) sinθ/cosθ + cosθ/sinθ = cscθsecθ

2) (tanB + cotB)/tanB=csc^2B

Follow
•
1

Add comment

More

Report

William W. answered • 12/12/20

Tutor

5.0
(730)
Experienced Tutor and Retired Engineer

Sam Z. answered • 12/12/20

Tutor

4.3
(12)
Math/Science Tutor

1) sinθ/cosθ + cosθ/sinθ = cscθsecθ

tanθ + cotθ = 1/(sinθ*cosθ)

1st quad; everything is=.

Denise G. answered • 12/12/20

Tutor

5.0
(405)
Algebra, College Algebra, Prealgebra, Precalculus, GED, ASVAB Tutor

For the first one, get a common denominator.

sinθ/cosθ(sinθ/sinθ) + cosθ/sinθ(cosθ/cosθ)

The result is:

(sin^{2}θ+cos^{2}θ)/(sinθcosθ)

Using the Pythagorean identities, the numerator is 1

The result is:

1/(sinθcosθ)

Using the reciprocal identities, you get the right side.

**cscθsecθ**

For the second one:

Substitute in all sin and cos for tan and cot

The result is:

(sinB/cosB + cosB/sinB)/(sinB/cosB)

The numerator is the same as problem 1 once you get a common denominator

[(sin^{2}B+cos^{2}B)/(sinBcosB)]/(sinB/cosB)

Use the Pythagorean identities again:

[1/(sinBcosB)]/(sinB/cosB)

Divide the fractions by multiplying by the reciprocal

[1/(sinBcosB)] x (cosB/sinB)

The cosB cancel out leaving 2 sins in the denominator

1/(sin^{2}B)

Your can simplify the 2 sins using the reciprocal function and that gives you the right side

**csc**^{2}**B**

Ask a question for free

Get a free answer to a quick problem.

Most questions answered within 4 hours.

Find an Online Tutor Now

Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.