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If
tan
−
1
(
b
c
+
a
)
+
tan
−
1
(
c
a
+
b
)
=
π
4
\tan^{-1}\left(\frac{b}{c+a}\right) + \tan^{-1}\left(\frac{c}{a+b}\right) = \frac{\pi}{4}
tan
−
1
(
c
+
a
b
)
+
tan
−
1
(
a
+
b
c
)
=
4
π
…
Mathematics
Problem
If
tan
−
1
(
b
c
+
a
)
+
tan
−
1
(
c
a
+
b
)
=
π
4
\tan^{-1}\left(\frac{b}{c+a}\right) + \tan^{-1}\left(\frac{c}{a+b}\right) = \frac{\pi}{4}
tan
−
1
(
c
+
a
b
)
+
tan
−
1
(
a
+
b
c
)
=
4
π
, where
a
,
b
,
c
a, b, c
a
,
b
,
c
are the sides of
Δ
A
B
C
\Delta ABC
Δ
A
B
C
, then
Δ
A
B
C
\Delta ABC
Δ
A
B
C
is a. Acute-angled triangle b. Obtuse-angled triangle c. Right-angled triangle d. Equilateral triangle
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