Cos alpha - beta by 2 = 2cos alpha + beta by 2, then find tan alpha by 2 into tan beta by 2
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Substitute the values of cosα and cosβ in terms of the tangents of their half angles.
cosα=
1+tan
2
2
α
1−tan
2
2
α
and cosβ=
1+tan
2
2
β
1−tan
2
2
β
Let tan
2
2
α
=a and tan
2
2
β
=b
Substituting the above values in the expression,
2(
1+b
1−b
−
1+a
1−a
)+(
1+a
1−a
)(
1+b
1−b
)=1
⇒2(
1+a+b+ab
1−b+a−ab−1+a−b+ab
)+
1+a+b+ab
1−a−b+ab
=1
⇒4(a−b)+1−a−b+ab=1+a+b+ab
⇒4a−4b=2b+2a
⇒2a=6b⇒a=3b
⇒tan
2
2
α
=3tan
2
2
β
⇒tan
2
2
α
−3tan
2
2
β
=0
⇒(tan
2
α
+
3
tan
2
β
)(tan
2
α
−
3
tan
2
β
)=0
∴tan
2
α
+
3
tan
2
β
=0 or tan
2
β
−
3
tan
2
β
=0
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