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Byju's Answer
Standard XII
Mathematics
Vn Method
Suppose α, β ...
Question
Suppose
α
,
β
and
θ
are angles satisfying
0
<
α
<
θ
<
β
<
π
2
, then
sin
α
−
sin
β
cos
β
−
cos
α
=
A
tan
θ
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B
−
tan
θ
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C
cot
θ
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D
−
cot
θ
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Solution
The correct option is
C
cot
θ
Let
f
(
x
)
=
sin
x
and
g
(
x
)
=
cos
x
, then
f
and
g
are continuous and derivable. Also,
sin
x
≠
0
for any
x
∈
(
0
,
π
2
)
So by Cauchy's mean value theorem,
f
(
β
)
−
f
(
α
)
g
(
β
)
−
g
(
α
)
=
f
′
(
θ
)
g
′
(
θ
)
⇒
sin
β
−
sin
α
cos
β
−
cos
α
=
cos
θ
−
sin
θ
=
−
cot
θ
⇒
sin
α
−
sin
β
cos
β
−
cos
α
=
cot
θ
Suggest Corrections
2
Similar questions
Q.
If
cos
(
θ
−
α
)
=
a
and
sin
(
θ
−
β
)
=
b
(
0
)
<
θ
−
α
,
θ
−
β
<
π
2
)
, then
cos
2
(
α
−
β
)
+
2
a
b
sin
(
α
−
β
)
is equal to
Q.
Consider the points
P
=
(
−
sin
(
β
−
α
)
,
−
cos
β
)
,
Q
=
(
cos
(
β
−
α
)
,
sin
β
)
and
R
=
(
cos
(
β
−
α
+
θ
)
,
sin
(
β
−
θ
)
)
, where
0
<
α
,
β
<
π
4
then
Q.
If
α
,
β
,
γ
are acute angles and
cos
θ
=
sin
β
sin
α
,
cos
ϕ
=
sin
γ
sin
α
and
cos
(
θ
−
ϕ
)
−
sin
β
sin
γ
=
0
, then
tan
2
α
−
tan
2
β
−
tan
2
γ
is equal to
Q.
Show that
cos
3
(
α
+
θ
)
sin
3
(
β
−
γ
)
+
cos
3
(
β
+
θ
)
sin
3
(
γ
−
α
)
+
cos
3
(
γ
+
θ
)
sin
3
(
α
−
β
)
=
3
cos
(
α
+
θ
)
cos
(
β
+
θ
)
cos
(
γ
+
θ
)
×
sin
(
α
−
β
)
sin
(
β
−
γ
)
sin
(
γ
−
α
)
Q.
If
cos
α
cos
θ
+
sin
α
sin
θ
=
cos
β
cos
θ
+
sin
β
sin
θ
=
1
where
α
and
β
do not differ by an even multiple of
π
, then show
cos
α
cos
β
cos
2
θ
+
sin
α
sin
β
sin
2
θ
+
1
=
0
.
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Vn Method
Standard XII Mathematics
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