Educator

Given that tanθ = - 5.3 and [(3π)/2] < θ < 2π, find secθ.

• Use the Pythagorean Identity to solve for secθ: 1 + tan²θ = sec²θ
• 1 + ( − 5.31)² = sec²θ ⇒ 1 + 28.1961 = sec²θ ⇒ sec²θ = 29.1961 ⇒ secθ = ±√{29.1961}
• secθ = ± 5.4 Remember the mnemonic ASTC. This tells which quadrant cosine values will be positive. Cosine is only positive in quadrants I and IV which means the same holds true for secant values.

secθ = 5.4 because θ is in quadrant IV

Given that secθ = [13/5] and 180^° < θ < 270^° , find tanθ.

• Use the Pythagorean Identity to solve for tanθ: 1 + tan²θ = sec²θ
• 1 + tan²θ = ( [13/5] )² ⇒ 1 + tan²θ = [169/25] ⇒ tan²θ = [169/25] - [25/25] ⇒ tanθ = ±√{[144/25]}
• tanθ = ±[12/5] Remember the mnemonic ASTC. This tells which quadrant tangent values will be positive. Tangent is only positive in quadrants I and III.

tanθ = [12/5] because θ is in quadrant III

Given that tanθ = - 3.23 and [(π)/2] < θ < π, find secθ.

• Use the Pythagorean Identity to solve for secθ: 1 + tan²θ = sec²θ
• 1 + ( − 3.23)² = sec²θ ⇒ 1 + 10.4329 = sec²θ ⇒ sec²θ = 11.4329 ⇒ secθ = ±√{11.4329}
• secθ = ± 3.38126 Remember the mnemonic ASTC. This tells which quadrant cosine values will be positive. Cosine is only positive in quadrants I and IV which means the same holds true for secant values.

secθ = - 3.4 because θ is in quadrant II

Given that secθ = − [5/4] and 90^° < θ < 180^°, find tanθ.

• Use the Pythagorean Identity to solve for tanθ: 1 + tan²θ = sec²θ
• 1 + tan²θ = ( − [5/4] )² ⇒ 1 + tan²θ = [25/16] ⇒ tan²θ = [25/16] - [16/16] ⇒ tanθ = ±√{[9/16]}
• tanθ = ±[3/4] Remember the mnemonic ASTC. This tells which quadrant tangent values will be positive. Tangent is only positive in quadrants I and III.

tanθ = − [3/4] because θ is in quadrant II

Verify the following identity: (tan²θ + 1)(cos²x - 1) = - tan²x

• Try to get the left hand side to look like the right hand side because it is the more complicated side
• (sec²θ)( - sin²x) = - tan²x by the Pythagorean Identities of sin²θ + cos²θ = 1 and 1 + tan²θ = sec²θ
• [1/(cos²x)] · - sin²x = - tan²x by the Reciprocal Identity
• [( − sin²x)/(cos²x)] = - tan²x by multiplying
• ( [( − sinx)/cosx] )² = - tan²x by the Rule of Exponents

- tan²x = - tan²x by the Quotient Identity

Verify the following identity: secθ + tanθ = [(cosθ)/(1 − sinθ)]

• Try to get the right hand side to look like the left hand side bcause it is the more complicated side
• secθ + tanθ = [(cosθ)/(1 − sinθ)] · [(1 + sinθ)/(1 + sinθ)], multiple by the conjugate
• secθ + tanθ = [(cosθ+ cosθsinθ)/(1 − sin²θ)], multiplication of fractions
• secθ + tanθ = [(cosθ+ cosθsinθ)/(cos²θ)], Pythagorean Identity sin²θ + cos²θ = 1
• secθ + tanθ = [(cosθ)/(cos²θ)] + [(cosθsinθ)/(cos²θ)], Separate Fractions
• secθ + tanθ = [1/(cosθ)] + [(sinθ)/(cosθ)], Simplifying

secθ + tanθ = secθ + tanθ, Reciprocal Identities

Verify the following identity: tanθ + cotθ = secθcscθ

• Try to get the left hand side to look like the right hand side bcause it is the more complicated side
• [(sinθ)/(cosθ)] + [(cosθ)/(sinθ)] = secθcscθ, Quotient Identity
• [(sin²θ+ cos²θ)/(cosθsinθ)] = secθcscθ, Adding Fractions
• [1/(cosθsinθ)] = secθcscθ, Pythagorean Identity
• [1/(cosθ)] · [1/(sinθ)] = secθcscθ, Product of Fractions

secθcscθ = secθcscθ, Reciprocal Identities

Verify the following identity: [(cot²θ)/(cscθ)] = cscθ - sinθ

• Try to get the left hand side to look like the right hand side bcause it is the more complicated side
• [(csc²θ− 1)/(cscθ)] = cscθ - sinθ, Pythagorean Identity: csc²θ - 1 = cot²θ
• [(csc²theta)/(cscθ)] - [1/(cscθ)] = cscθ - sinθ, Separate Fractions
• cscθ - [1/(cscθ)] = cscθ - sinθ, Simplifying

cscθ - sinθ = cscθ - sinθ, Reciprocal Identity

Verify the following identity: cot²θ(sec²θ - 1) = 1

• Try to get the left hand side to look like the right hand side bcause it is the more complicated side
• cot²θtan²θ = 1, Pythagorean Identityq: tan²θ = sec²θ - 1
• ( [(cos²θ)/(sin²θ)] ) ( [(sin²θ)/(cos²θ)] ) = 1, Reciprocal Identities

1 = 1, Simplifying

Verify the following identity: [(cot²θ)/(1 + cscθ)] = [(1 − sinθ)/(sinθ)]

• [(csc²θ− 1)/(1 + cscθ)] = [(1 − sinθ)/(sinθ)], Pythagorean Identity on the left hand side
• [((cscθ− 1)(cscθ+ 1))/(1 + cscθ)] = [(1 − sinθ)/(sinθ)], Factor the left hand side
• cscθ - 1 = [(1 − sinθ)/(sinθ)], Simplify the left hand side
• cscθ - 1 = [1/(sinθ)] - [(sinθ)/(sinθ)], Separate Fractions on the right hand side

cscθ - 1 = cscθ - 1, Reciprocal Identity on the right hand side

1. Intro
2. Inverse Trigonometric Function Domains and Ranges
• Arcsine
• Arccosine
• Arctangent
3. Example 1: Arcsines of Common Values
4. Example 2: Odd, Even, or Neither
5. Example 3: Arccosines of Common Values
6. Extra Example 1: Arctangents of Common Values
7. Extra Example 2: Arcsin/Arccos/Arctan Values