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Tardigrade
Question
Mathematics
Assertion: | cos (θ+α) cos (θ+β) cos (θ+γ) sin (θ+α) sin (θ+β) sin (θ+γ) sin (β-γ) sin (γ-α) sin (α-β)| is independent of θ Reason: If f(θ)=c, then f(θ) is independent of θ.
Q.
Assertion:
∣
∣
cos
(
θ
+
α
)
sin
(
θ
+
α
)
sin
(
β
−
γ
)
cos
(
θ
+
β
)
sin
(
θ
+
β
)
sin
(
γ
−
α
)
cos
(
θ
+
γ
)
sin
(
θ
+
γ
)
sin
(
α
−
β
)
∣
∣
is independent of
θ
Reason:
If
f
(
θ
)
=
c
, then
f
(
θ
)
is independent of
θ
.
2109
196
Determinants
Report Error
A
Assertion is correct, reason is correct; reason is a correct explanation for assertion.
100%
B
Assertion is correct, reason is correct; reason is not a correct explanation for assertion
0%
C
Assertion is correct, reason is incorrect
0%
D
Assertion is incorrect, reason is correct.
0%
Solution:
Let
f
(
θ
)
=
∣
∣
cos
(
θ
+
α
)
sin
(
θ
+
α
)
sin
(
β
−
γ
)
cos
(
θ
+
β
)
sin
(
θ
+
β
)
sin
(
γ
−
α
)
cos
(
θ
+
γ
)
sin
(
θ
+
γ
)
sin
(
α
−
β
)
∣
∣
∴
f
′
(
θ
)
=
∣
∣
−
sin
(
θ
+
α
)
sin
(
θ
+
α
)
sin
(
β
−
γ
)
−
sin
(
θ
+
β
)
sin
(
θ
+
β
)
sin
(
γ
−
α
)
−
sin
(
θ
+
γ
)
sin
(
θ
+
γ
)
sin
(
α
−
β
)
∣
∣
+
∣
∣
cos
(
θ
+
α
)
cos
(
θ
+
α
)
sin
(
β
−
γ
)
cos
(
θ
+
β
)
cos
(
θ
+
β
)
sin
(
γ
−
α
)
cos
(
θ
+
γ
)
cos
(
θ
+
γ
)
sin
(
α
−
β
)
∣
∣
+
∣
∣
cos
(
θ
+
α
)
sin
(
θ
+
α
)
0
cos
(
θ
+
β
)
sin
(
θ
+
β
)
0
cos
(
θ
+
γ
)
sin
(
θ
+
γ
)
0
∣
∣
=
0
+
0
+
0
=
0
⇒
f
′
(
θ
)
=
0
⇒
f
(
θ
)
=
c