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Answers
Answer:
1
+
cot
2
θ
=
csc
2
θ
is found by rewriting the left side of the equation in terms of sine and cosine.
Prove:
1
+
cot
2
θ
=
csc
2
θ
1
+
cot
2
θ
=
(
1
+
cos
2
θ
sin
2
θ
)
Rewrite the left side
.
=
(
sin
2
θ
sin
2
θ
)
+
(
cos
2
θ
sin
2
θ
)
Write both terms with the common denominator
.
=
sin
2
θ
+
cos
2
θ
sin
2
θ
=
1
sin
2
θ
=
csc
2
θ
Similarly,
1
+
tan
2
θ
=
sec
2
θ
can be obtained by rewriting the left side of this identity in terms of sine and cosine. This gives
1
+
tan
2
θ
=
1
+
(
sin
θ
cos
θ
)
2
Rewrite left side
.
=
(
cos
θ
cos
θ
)
2
+
(
sin
θ
cos
θ
)
2
Write both terms with the common denominator
.
=
cos
2
θ
+
sin
2
θ
cos
2
θ
=
1
cos
2
θ
=
sec
2
θ
The next set of fundamental identities is the set of even-odd identities. The even-odd identities relate the value of a trigonometric function at a given angle to the value of the function at the opposite angle and determine whether the ident
Answer:
3 tan³ θ = tan θ
Divide each side of the equation by tan θ
3 * tan³ θ tan θ
-------------- = ---------
tan θ tan θ
3 * tan³ θ tan θ
-------------- = -------------
tan θ tan θ
3 * tan³ θ
-------------- = 1
tan θ
3 * tan θ * tan² θ
----------------------- = 1
tan θ
3 * tan θ * tan² θ
----------------------- = 1
tan θ
3 tan² θ = 1
Divide each side of the equation by 3
3 * tan² θ 1
-------------- = -----
3 3
3 * tan² θ 1
-------------- = ------
3 3
1
(1) * tan² θ = -----
3
1
tan² θ = -----
3
Take the square root of each side of the equation
1
√(tan² θ) = √(-----)
3
1
tan θ = ± √(-----)
3
√1
tan θ = ± ------
√3
1
tan θ = ± ------
√3
To rationalize the denominator, multiply
the right side of the equation by √3/√3
1 √3
tan θ = ± ----- * ------
√3 √3
1 * √3
tan θ = ± -------------
√3 * √3
√3
tan θ = ± -------
3
θ = tan⁻¹[±√(3) / 3]
θ = tan⁻¹[±1.7320508075689 / 3]
θ = tan⁻¹[±0.5773502691896]
θ = 30°, 150°, 210°, 330°
(or θ = 1pi/6 radians, 5pi/6 radians,
7pi/6 radians, 11pi/6 radians)
θ =180k + 30
θ =180k - 30