prove that √3 cot 20 × cot 40 × cot 80 = 1
Answers
Answer:
Step-by-step explanation:
Formula used:
Step-by-step explanation:
∗cot20∗cot40∗cot80=1
Step-by-step explanation:
Formula used:
tan(A-B)=\frac{tanA-tanB}{1+tanA\:tanB}tan(A−B)=
1+tanAtanB
tanA−tanB
tan(A+B)=\frac{tanA+tanB}{1-tanA\:tanB}tan(A+B)=
1−tanAtanB
tanA+tanB
tan3A=\frac{3tanA-tan^3A}{1-3tan^2A}tan3A=
1−3tan
2
A
3tanA−tan
3
A
\sqrt3*cot20*cot40*cot80
3
∗cot20∗cot40∗cot80
=\sqrt3*(\frac{1}{tan20})*(\frac{1}{tan40})*tan10=
3
∗(
tan20
1
)∗(
tan40
1
)∗tan10
=\sqrt3*(\frac{1}{tan(30-10)})*(\frac{1}{tan(30+10)})*tan10=
3
∗(
tan(30−10)
1
)∗(
tan(30+10)
1
)∗tan10
=\sqrt3*(\frac{1+tan30\:tan10}{tan30-tan10})*(\frac{1-tan30\:tan10}{tan30+tan10})*tan10=
3
∗(
tan30−tan10
1+tan30tan10
)∗(
tan30+tan10
1−tan30tan10
)∗tan10
=\sqrt3*(\frac{1^2-tan^230\:tan^210}{tan^230-tan^210})*tan10=
3
∗(
tan
2
30−tan
2
10
1
2
−tan
2
30tan
2
10
)∗tan10
=\sqrt3*(\frac{1-(\frac{1}{\sqrt3})^2tan^210}{(\frac{1}{\sqrt3})^2-tan^210})*tan10=
3
∗(
(
3
1
)
2
−tan
2
10
1−(
3
1
)
2
tan
2
10
)∗tan10
=\sqrt3*(\frac{1-(\frac{1}{3})tan^210}{(\frac{1}{3})-tan^210})*tan10=
3
∗(
(
3
1
)−tan
2
10
1−(
3
1
)tan
2
10
)∗tan10
=\sqrt3*(\frac{3-tan^210}{1-3tan^210})*tan10=
3
∗(
1−3tan
2
10
3−tan
2
10
)∗tan10
=\sqrt3*(\frac{3tan10-tan^310}{1-3tan^210})=
3
∗(
1−3tan
2
10
3tan10−tan
3
10
)
=\sqrt3*tan3(10)=
3
∗tan3(10)
=\sqrt3*tan30=
3
∗tan30
=\sqrt3*\frac{1}{\sqrt3}=
3
∗
3
1
=1=1