if 1/under root5 +2 + 1/2+under root3 +1/under root3+ under root2+ 1/under root2+1 =a+b under root c then find the value of a+b+c
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Answer:
cot7
2
1
°
=
2
+
3
+
4
+
6
⇒ L.H.S= cot7
2
1
°
= cot
2
15
°
=
sin
(
2
15
)
°
cos
(
2
15
)
°
=
2 sin
(
2
15
)
°
2 cos
(
2
15
)
°
=
2sin
(
2
15
)
°
cos
(
2
15
)
°
2cos
2
(
2
15
)
°
=
sin15°
1+cos15°
[∵1+cos2θ=2cos
2
2
2θ
]
[&2sin
2
θ
.cos
2
θ
=sin
2
2θ
]
⇒ cos15°=
4
1
(
6
+
2
)
&sin15°=
4
1
(
6
−
2
)
∴ LHS=
4
1
(
6
−
2
)
1+
4
1
(
6
+
2
)
=
6
−
2
4+
6
+
2
Apply componendo and dividendo, we get
⇒LHS=
(
6
−
2
)
(
6
+
2
)
(4+
6
+
2
)
(
6
+
2
)
=
6−2
4
6
+4
2
+6+
12
+
12
+2
=
4
4
6
+4
2
+8+2
3
+2
3
=
4
8
+
4
4
2
+
4
4
3
+
4
4
6
= 2+
2
+
3
+
6
=
2
+
3
+
4
+
6
⇒ R.H.S
Hence, the answer is prove.
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