tan (180+theta) = ?
.
Answers
Step-by-step explanation:
Trigonometric ratios of 180 plus theta (180° + θ).
Let the ray OA revolves from its initial position of OX and makes an angle of
Θ
i.e.
a
n
g
l
e
X
O
A
=
Θ
. Let us consider another ray OB which rotates and makes an angle (180 +
Θ
) i.e.
a
n
g
l
e
X
O
B
=
180
+
Θ
. Let P(x,y) and Q(-x,-y) on ray OA and OB respectively in such a way that OA = OB = r. Now draw a perpendicular from P and Q on ray OA and OB respectively. Let the foot of the perpendicular be M and N.
Now as the triangle OPM and OQN are right triangles so they are congruent by HL (RHS) theorem.
OM = ON = x and PM and QN = y
⇒ coordinates of Q are (-x,-y)
trigonometric ratios of (180 +Θ)
trigonometric ratios of (180 +Θ)
0Save
Trigonometric ratios are :
s
i
n
(
180
+
Θ
)
=
Q
N
O
Q
=
−
y
r
But
s
i
n
Θ
=
P
M
O
P
( OP = OQ and PM = QN)
∴
s
i
n
(
180
+
Θ
)
=
−
s
i
n
Θ
–––––––––––––––––––––––––
c
o
s
(
180
+
Θ
)
=
O
N
O
Q
=
−
x
r
But
c
o
s
Θ
=
O
M
O
P
( OP = OQ and OM = ON)
∴
c
o
s
(
180
+
Θ
)
=
−
c
o
s
Θ
–––––––––––––––––––––––––
t
a
n
(
180
+
Θ
)
=
Q
N
O
N
=
−
y
−
x
But
t
a
n
Θ
=
P
M
O
M
( QN = PM and OM = ON)
∴
t
a
n
(
180
+
Θ
)
=
t
a
n
Θ
––––––––––––––––––––––––
As we know that,
c
s
c
Θ
=
1
s
i
n
Θ
∴
c
s
c
(
180
+
Θ
)
=
1
s
i
n
(
180
+
Θ
)
But
s
i
n
(
180
+
Θ
)
=
−
s
i
n
Θ
∴
c
s
c
(
180
+
Θ
)
=
1
−
s
i
n
Θ
)
c
s
c
(
180
+
Θ
)
=
−
c
s
c
Θ
–––––––––––––––––––––––––
Again,
s
e
c
Θ
=
1
c
o
s
Θ
∴
s
e
c
(
180
+
Θ
)
=
1
c
o
s
(
180
+
Θ
)
But
c
o
s
(
180
+
Θ
)
=
−
c
o
s
Θ
∴
s
e
c
(
180
+
Θ
)
=
1
−
c
o
s
Θ
)
s
e
c
(
180
+
Θ
)
=
−
s
e
c
Θ
–––––––––––––––––––––––––
Now,
c
o
t
Θ
=
1
t
a
n
Θ
∴
c
o
t
(
180
+
Θ
)
=
1
t
a
n
(
180
+
Θ
)
But
t
a
n
(
180
+
Θ
)
=
t
a
n
Θ
∴
c
o
t
(
180
+
Θ
)
=
1
t
a
n
Θ
)
c
o
t
(
180
+
Θ
)
=
c
o
t
Θ
–––––––––––––––––––––––