The value of tan 10° . tan 30° . tan 50° . tan 70°
Answers
Answer:
Answer:
\tan 10\tan 50\tan 70=\tan 30tan10tan50tan70=tan30
Step-by-step explanation:
We need to prove that
\tan 10\tan 50\tan 70=\tan 30tan10tan50tan70=tan30
Taking LHS,
LHS=\tan 10\tan 50\tan 70LHS=tan10tan50tan70
It can be rewritten as
LHS=\tan 10\tan (60-10)\tan (60+10)LHS=tan10tan(60−10)tan(60+10)
We know that
\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}tan(A+B)=
1−tanAtanB
tanA+tanB
and \tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}tan(A−B)=
1+tanAtanB
tanA−tanB
Using the above formulas we get
LHS=\tan 10(\frac{\tan 60-\tan 10}{1+\tan 60\tan 10})(\frac{\tan 60+\tan 10}{1-\tan 60\tan 10})LHS=tan10(
1+tan60tan10
tan60−tan10
)(
1−tan60tan10
tan60+tan10
)
LHS=\tan 10(\frac{\tan^2 60-\tan^2 10}{1^2-(\tan 60\tan 10)^2})LHS=tan10(
1
2
−(tan60tan10)
2
tan
2
60−tan
2
10
)
LHS=\tan 10(\frac{(\sqrt 3)^2-\tan^2 10}{1-(\sqrt 3)^2\tan^2 10})LHS=tan10(
1−(
3
)
2
tan
2
10
(
3
)
2
−tan
2
10
) [\because a^2-b^2=(a-b)(a+b)][∵a
2
−b
2
=(a−b)(a+b)]
LHS=\tan 10(\frac{3-\tan^2 10}{1-3\tan^2 10})LHS=tan10(
1−3tan
2
10
3−tan
2
10
) [\because \tan 60=\sqrt{3}][∵tan60=
3
]
LHS=\frac{3\tan 10-\tan^3 10}{1-3\tan^2 10}LHS=
1−3tan
2
10
3tan10−tan
3
10
LHS=\tan (3\times 10)LHS=tan(3×10) \tan 3x=\frac{3\tan x-\tan^3x}{1-3\tan^2x}tan3x=
1−3tan
2
x
3tanx−tan
3
x
LHS=\tan 30LHS=tan30
LHS=RHSLHS=RHS
Hence proved.