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Other
Quantitative Aptitude
Functions
If tanα-tan...
Question
If
tan
α
−
tan
β
=
m
and
cot
α
−
cot
β
=
n
, then prove that
cot
(
α
−
β
)
=
1
m
−
1
n
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Solution
tan
(
α
−
β
)
=
tan
α
−
tan
β
1
+
tan
α
tan
β
cot
(
α
−
β
)
=
1
+
tan
α
tan
β
tan
α
−
tan
β
.
.
.
.
(
1
)
cot
(
α
−
β
)
=
1
tan
α
−
tan
β
+
tan
α
tan
β
tan
α
−
tan
β
cot
α
−
cot
β
=
n
1
tan
α
−
1
tan
β
=
n
⇒
1
+
tan
α
tan
β
tan
α
−
tan
β
=
−
1
n
.
.
.
.
(
2
)
From equation (1) and (2)
cot
(
α
−
β
)
=
1
m
−
1
n
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