Prove the following identity:1)tan A= n tan B and sin A= m sin B.Prove that cos²A=m²-1/n²-1
Answers
Answer:
Step-by-step explanation:
Given that,
tan
A
=
n
tan
B
⇒
sin
A
cos
A
=
n
⋅
sin
B
cos
B
.
⇒
sin
A
sin
B
=
n
⋅
cos
A
cos
B
...
...
...
...
...
...
...
...
...
...
...
...
.
⟨
1
⟩
.
Also given that,
sin
A
=
m
sin
B
⇒
sin
A
sin
B
=
m
...
...
...
...
...
...
...
⟨
2
⟩
.
Comparing
⟨
1
⟩
and
⟨
2
⟩
,
we get,
m
=
n
⋅
cos
A
cos
B
,
giving,
cos
B
=
n
m
⋅
cos
A
...
...
⟨
3
⟩
.
⟨
2
⟩
⇒
sin
B
=
1
m
⋅
sin
A
...
...
...
...
...
...
...
...
...
...
...
.
.
⟨
2
'
⟩
.
Now, using
⟨
2
'
⟩
and
⟨
3
⟩
in
cos
2
B
+
sin
2
B
=
1
,
we get,
⇒
n
2
m
2
⋅
cos
2
A
+
1
m
2
⋅
sin
2
A
=
1
.
⇒
n
2
cos
2
A
+
sin
2
A
=
m
2
.
⇒
n
2
cos
2
A
+
(
1
−
cos
2
A
)
=
m
2
.
⇒
n
2
cos
2
A
−
cos
2
A
=
m
2
−
1
,
i
.
e
.
,
⇒
(
n
2
−
1
)
cos
2
A
=
(
m
2
−
1
)
.
⇒
cos
2
A
=
m
2
−
1
n
2
−
1
,
as desired!
Enjoy Maths.!
Formula used:
tan²a-tan²b = (sin²a-sin²b)/cos²a*cos²b
