62nd Problem 2016

Geometry Level 2

tan ⁡ 2 θ = ? \tan ^{ 2 }{ \theta =? }

Check out the set: 2016 Problems

csc ⁡ 2 θ tan ⁡ 2 θ − 1 \csc ^{ 2 }{ \theta } \tan ^{ 2 }{ \theta } - 1 sin ⁡ 2 θ tan ⁡ 2 θ − 1 \sin ^{ 2 }{ \theta } \tan ^{ 2 }{ \theta } -1 cos ⁡ 2 θ tan ⁡ 2 θ − 1 \cos ^{ 2 }{ \theta } \tan ^{ 2 }{ \theta } - 1 cos ⁡ 2 θ cot ⁡ 2 θ − 1 \cos ^{ 2 }{ \theta } \cot ^{ 2 }{ \theta } - 1

Angela Fajardo by Angela Fajardo , 19, Philippines

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1 solution

Ashish Menon
Mar 23, 2016

csc ⁡ 2 θ tan ⁡ 2 θ − 1 \csc ^2\theta \tan^2\theta -1

= ( 1 sin ⁡ 2 θ × sin ⁡ 2 θ cos ⁡ 2 θ ) − 1 (\dfrac {1}{\sin^2\theta} × \dfrac {\sin^2\theta}{\cos^2\theta}) - 1

= 1 cos ⁡ 2 θ − 1 \dfrac {1}{\cos^2\theta} - 1

= 1 − cos ⁡ 2 θ cos ⁡ 2 θ \dfrac {1- \cos^2\theta}{\cos^2\theta}

= sin ⁡ 2 θ cos ⁡ 2 θ \dfrac {\sin^2\theta}{\cos^2\theta}

= tan ⁡ 2 θ □ \boxed {\tan^2\theta} _\square

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Moderator note:

Simple standard approach.

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