1.
3(sin 0 + cos 0) – (sin 0 + cos 0)^3 = 2(sin^3 0 + cos^3 0)
Answers
Answer:
Convert 8cosec
3
(2θ) into 8×
sin
3
(2θ)
1
=8×
(2sinθcosθ)
3
1
=
sin
3
θcos
3
θ
1
Similarly we write tan
3
θ and cot
3
θ in terms of sinθ and cosθ.
Denote sinθ=S and cosθ=C
So,
C
3
S
3
+
S
3
C
3
=12+
S
3
C
3
1
Take L.C.M, we get
S
3
C
3
S
6
+C
6
=
S
3
C
3
12S
3
C
3
+1
Now we cancel S
3
C
3
but also note that means S=0 and C=0 can't be solution of the equation because they are in denominator.
⇒sinθ
=0,cosθ
=0
So equation now is, S
6
+C
6
=12S
3
C
3
+1
We know S
6
+C
6
=1−3S
2
C
2
.
Placing in above equation,
1−3S
2
C
2
=12S
3
C
3
+1→3S
2
C
2
(4SC+1)=0
S
=0,C
=0→4SC+1=0⇒4sinθcosθ+1=0⇒2sin2θ+1=0
sin2θ=−
2
1
⇒2θ=(2n+1)π+
6
π
and 2θ=(2n+1)π+
6
5π
θ=(2n+1)
2
π
+
12
π
and θ=(2n+1)
2
π
+
12
5π
Hence put n=0,1 we get θ=
12
7π
,
12
11π
,
12
19π
,
12
23π
Hence, (A)(B)(C)(D)
Step-by-step explanation:
I hope it's helpful