trigonometry identity
Answers
Step-by-step explanation:
Trigonometric Identity 1
Now, divide each term of equation (1) by AC2, we have
A
C
2
A
C
2
=
A
B
2
A
C
2
+
B
C
2
A
C
2
⇒
A
B
2
A
C
2
+
B
C
2
A
C
2
=
1
⇒
(
A
B
A
C
)
2
+
(
B
C
A
C
)
2
=
1
………………………….(2)
We know that,
(
A
B
A
C
)
2
=
cos
a
and
(
B
C
A
C
)
2
=
sin
a
, thus equation (2) can be written as-
sin2 a + cos2 a = 1
Identity 1 is valid for angles 0 ≤ a ≤ 90.
Trigonometric Identity 2
Now Dividing the equation (1) by AB2, we get
A
C
2
A
B
2
=
A
B
2
A
B
2
+
B
C
2
A
B
2
⇒
A
C
2
A
B
2
=
1
+
B
C
2
A
B
2
⇒
(
A
C
A
B
)
2
=
1
+
(
B
C
A
B
)
2
……………………(3)
By referring trigonometric ratios, it can be seen that:
A
C
A
B
=
h
y
p
o
t
e
n
u
s
e
s
i
d
e
a
d
j
a
c
e
n
t
t
o
a
n
g
l
e
a
=
sec
a
Similarly,
B
C
A
B
=
s
i
d
e
o
p
p
o
s
i
t
e
t
o
a
n
g
l
e
a
s
i
d
e
a
d
j
a
c
e
n
t
t
o
a
n
g
l
e
a
=
tan
a
Replacing the values of
A
C
A
B
and
B
C
A
B
in the equation (3) gives,
1+tan2 a = sec2 a
As it is known that tan a is not defined for a = 90° therefore identity 2 obtained above is true for 0 ≤ A <90.
Trigonometric Identity 3
Dividing the equation (1) by BC2, we get
A
C
2
B
C
2
=
A
B
2
B
C
2
+
B
C
2
B
C
2
⇒
A
C
2
B
C
2
=
A
B
2
B
C
2
+
1
⇒
(
A
C
B
C
)
2
=
(
A
B
B
C
)
2
+
1
…………………..(iv)
By referring trigonometric ratios, it can be seen that:
A
C
B
C
=
h
y
p
o
t
e
n
u
s
e
s
i
d
e
o
p
p
o
s
i
t
e
t
o
a
n
g
l
e
a
=
c
o
s
e
c
a
Also,
A
B
B
C
=
s
i
d
e
a
d
j
a
c
e
n
t
t
o
a
n
g
l
e
a
s
i
d
e
o
p
p
o
s
i
t
e
t
o
a
n
g
l
e
a
=
cot
a
Replacing the values of
A
C
B
C
and
A
B
B
C
in the equation (4) gives,
cosec2 a = 1 + cot2 a
Since cosec a and cot a are not defined for a = 0°, therefore the identity 3 is obtained is true for all the values of ‘a’ except at a = 0°. Therefore, the identity is true for all such that, 0° < a ≤ 90°.