One step answer

Geometry Level 2

tan ⁡ 4 0 ∘ + 2 tan ⁡ 1 0 ∘ = ? \large{\tan40^{\circ}+2\tan10^{\circ}} = \ ?

cot ⁡ 4 0 ∘ \cot40^{\circ} None of these choices cot ⁡ 6 0 ∘ \cot60^{\circ} cot ⁡ 5 0 ∘ \cot50^{\circ} cot ⁡ 1 0 ∘ \cot10^{\circ}

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Tanishq Varshney by Tanishq Varshney , 23, India

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3 solutions

Tanishq Varshney
Apr 30, 2015

tan ⁡ 4 0 ∘ + 2 cot ⁡ 8 0 ∘ \tan 40^{\circ}+2\cot 80^{\circ}

2 cot ⁡ 2 θ = cot ⁡ θ − tan ⁡ θ 2\cot 2 \theta=\cot \theta-\tan \theta

= cot ⁡ 4 0 ∘ \cot 40^{\circ}

24 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

Short and sweet! Great job.

How did you get 2 cot ⁡ 2 θ = cot ⁡ θ − tan ⁡ θ 2\cot{2\theta}=\cot{\theta}-\tan{\theta} ?

Thomas James Bautista - 6 years, 1 month ago

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ok u want me to prove this

2 t a n 2 θ \frac{2}{tan2 \theta}

t a n 2 θ = 2 t a n θ 1 − t a n 2 θ tan 2\theta=\frac{2tan\theta}{1-tan^{2} \theta}

putting the value of t a n 2 θ tan 2\theta

we get 1 − t a n 2 θ t a n θ \frac{1-tan^{2} \theta}{tan \theta}

1 t a n θ − t a n θ \frac{1}{tan \theta}-tan \theta

c o t θ − t a n θ cot \theta-tan \theta

hope it helps @Thomas James Bautista

Tanishq Varshney - 6 years, 1 month ago

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Thanks for that! After I commented, I tried proving it myself in a different way, but your proof works as well. I wasn't aware of this identity before, so I was quick to comment.

Thomas James Bautista - 6 years, 1 month ago

Another approach!! tan ⁡ 4 0 ∘ + 2 tan ⁡ 1 0 ∘ = tan ⁡ 4 0 ∘ + 2 tan ⁡ 8 0 ∘ = tan ⁡ 4 0 ∘ + 2 ∗ 1 − ( tan ⁡ 4 0 ∘ ) 2 2 ∗ tan ⁡ 4 0 ∘ = 1 tan ⁡ 4 0 ∘ = cot ⁡ 4 0 o \tan40^{\circ}+2\tan10^{\circ} =\tan40^{\circ}+\dfrac{2}{\tan80^{\circ}} \\=\tan40^{\circ}+2*\dfrac{1-(\tan40^{\circ})^2}{2*\tan40^{\circ}} =\dfrac{1}{\tan40^{\circ}}\\=\cot40^o

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

I like how double angle identities sneaks up in here. Good job!

Appan Rakaraddi
May 1, 2015

Tan40 + tan10 = tan50(1-tan10tan40)

Tan40+ tan10+tan10=tan50(1-tan10tan40)+tan10

=tan50

=cot40

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

Right, the conventional compound angle formula works too. Note that for clarity, at the second line, you have tan ⁡ ( 5 0 ∘ ) + tan ⁡ ( 1 0 ∘ ) ( 1 − tan ⁡ ( 5 0 ∘ ) tan ⁡ ( 4 0 ∘ ) ) \tan(50^\circ) + \tan(10^\circ) \left ( 1 - \tan(50^\circ) \tan(40^\circ) \right ) with the bracket equals to 0 0 .

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