SME
Grade 11 Math · Problem type
From tangent to secant
Use this for “tan θ = −12/5 in quadrant IV. Find sec θ and cos θ.”
Diagram
θ = 292.62°: the point (cos θ, sin θ) is ≈ (0.3846, −0.9231).
Reference angle: 67.38°.
tan
θ
=
sin θ
cos θ
≈
−2.4.
Quadrant: 4
Secant, sec θ: 13/5
This is an example. Type your own values to start a new problem.
(
)
2
+
1
=
13
5
2
Q
Quadrant
example
c
Cosine, cos θ
worked out (not typed)
s
Sine, sin θ
worked out (not typed)
θ
Angle
worked out (not typed) · °
Formulas
S
=
±
√
1 +
m
2
,
the
sign
from
quadrant
Q
13
5
=
±√(1
+
(−
12
5
)
2
),
the
sign
from
quadrant
4
c
=
1
S
5
13
=
1
13
5
s
=
m
×
c
−
12
13
=
−
12
5
×
5
13
θ
=
tan⁻¹(
m
),
moved
into
quadrant
Q
292.6199
=
tan⁻¹(−
12
5
),
moved
into
quadrant
4
Assumptions
•
Divide sin²θ + cos²θ = 1 by cos²θ: tan²θ + 1 = sec²θ.
•
sec θ = 1 ÷ cos θ has the sign of cos θ: positive in quadrants I and IV.
•
Then cos θ = 1 ÷ sec θ and sin θ = tan θ × cos θ.
Step-by-step
WE KNOW
m = −12/5, Q = 4
FIND
secant, sec θ (S), cosine, cos θ (c), sine, sin θ (s), angle (θ)
Step 1 · Find secant, sec θ (S)
Use S = ±√(1 + m²), the sign from quadrant Q
sec²
θ
=
1
+
tan²
θ
;
secant
is
positive
in
this
quadrant.
S
=
√
1 +
m
2
S
=
√(1
+
(−
12
5
)
2
)
S
=
√
1 +
144
25
S
=
√
169
25
S
=
13
5
Step 2 · Find cosine, cos θ (c)
Use c = 1 ÷ S
Cosine is the reciprocal of secant.
c
=
1
13
5
c
=
5
13
Step 3 · Find sine, sin θ (s)
Use s = m × c
tan
θ
=
sin
θ
cos
θ
,
so
multiply
by
cos
θ
.
s
=
−
12
5
×
5
13
s
=
−
12
13
Step 4 · Find angle (θ)
Use θ = tan⁻¹(m), moved into quadrant Q
tan⁻¹ gives a negative angle; a full turn on lands in quadrant IV.
θ
=
360
+
tan⁻¹(
m
)
θ
=
360
+
tan⁻¹(−
12
5
)
θ
=
360°
−
67.3801°
θ
=
292.6199°
Check
Put the numbers back in. Both sides should match.
13
5
=
√(1
+
(−
12
5
)
2
)
✓
5
13
=
1
13
5
✓
−
12
13
=
−
12
5
×
5
13
✓
292.6199
=
360
+
tan⁻¹(−
12
5
)
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.