A geometry problem by Elan Fenatic

Geometry Level 3

tan 9 tan 2 7 tan 6 3 + tan 8 1 = ? \tan9^\circ - \tan27^\circ - \tan63^\circ + \tan81^\circ = \, ?

4.4 2 4 6

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2 solutions

Rishabh Jain
Jul 6, 2016

( tan 8 1 + tan 9 ) ( tan 6 3 + tan 2 7 ) (\tan 81^\circ+\tan9^\circ)-(\tan 63^\circ+\tan27^\circ)

= ( sin 9 0 1 cos 9 cos 8 1 sin 9 sin 1 8 2 ) ( sin 9 0 1 cos 2 7 cos 6 3 sin 2 7 sin 5 4 2 ) =\left(\dfrac{\overbrace{\sin 90^{\circ}}^{\color{#D61F06}{1}}}{\underbrace{\cos 9^{\circ}\underbrace{\cos 81^{\circ}}_{\sin9^{\circ}}}_{\color{#D61F06}{\dfrac{\sin 18^{\circ}}{2}}}}\right)-\left( \dfrac{\overbrace{\sin 90^{\circ}}^{\color{#D61F06}{1}}}{\underbrace{\cos 27^{\circ}\underbrace{\cos 63^{\circ}}_{\sin 27^{\circ}}}_{\color{#D61F06}{\dfrac{\sin 54^{\circ}}2}}}\right)

= 2 ( 1 5 1 4 1 5 + 1 4 ) \large=2\left(\dfrac{1}{\frac{\sqrt 5-1}{4}}-\dfrac{1}{\frac{\sqrt 5+1}{4}}\right)

= 2 × 2 = 4 \huge =2\times 2=\boxed{\color{#3D99F6}{4}}


In second line I used : tan A + tan B = sin ( A + B ) cos A cos B , sin A cos A = sin 2 A 2 \small{\color{teal}{\tan A+\tan B=\dfrac{\sin (A+B)}{\cos A\cos B}~~,\sin A\cos A=\dfrac{\sin 2A}2}}

Also, sin 1 8 = 5 1 4 , sin 5 4 = 5 + 1 4 \small{\sin 18^{\circ}=\dfrac{\sqrt 5-1}{4},\sin 54^{\circ}=\dfrac{\sqrt 5+1}{4}}

Why don't you report this...Same question? You copied from there. :p

A Former Brilliant Member - 4 years, 11 months ago

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I just copied my solution... which I don't think must be an offense( I was not ready to type that thing once again :-P) . Maybe you should notify the problem poser and if I may tell you this is very common seeing question repeated on brilliant.. :-)

Rishabh Jain - 4 years, 11 months ago
Naitik Sanghavi
Jul 7, 2016

TanA + CotA=2Cosec2A

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