Prove that : tan a tan (a + 60°) + tan o tan (a - 60°) + tan (a + 60°) tan (a. -60=-3.
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Answer:
u tanθ+tan(60
o
+θ)+tan(120
o
+θ)=3tan3θ
LHS=tanθ+tan(60
o
+θ)+tan(120
o
+θ)
=tanθ+
(1−tan60
o
.tanθ)
(tan60
o
+tanθ)
+
(1−tan120
o
.tanθ)
(tan120
o
+tanθ)
=tanθ+
[1−
3
tanθ]
[
3
+tanθ]
+
[1+
3
tanθ]
[tanθ−
3
]
=
[1−3tan
2
θ]
tanθ+[
3
+3tanθ+tanθ+
3
tan
2
θ+tanθ−
3
−
3
tan
2
θ+ 3tanθ]
=tanθ+
[1−3tan
2
θ]
[8tanθ]
=
[1−3tan
2
θ]
[9tanθ−3tan
3
θ]
=3tan3θ
=RHS
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