Tangent Untangented

Geometry Level 2

Compute in degrees :

arctan ⁡ ( tan ⁡ 6 5 ∘ − 2 tan ⁡ 4 0 ∘ ) . \large \arctan( \tan 65^{\circ}- 2 \tan 40^{\circ}).


The answer is 25.

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Anish Harsha by Anish Harsha , 20, India

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4 solutions

Rishabh Jain
Jan 8, 2016

tan ⁡ 65 = tan ⁡ ( 45 + 20 ) = 1 + t a n 20 1 − t a n 20 \color{#3D99F6}{ \tan 65 = \tan (45+20) = \frac{1 + tan20}{1-tan20}} a n d , 2 tan ⁡ 40 = 2 ( 2 tan ⁡ 20 ) 1 − tan ⁡ 2 20 \color{#3D99F6}{ and,\space 2 \tan 40 = \frac{2(2 \tan 20)}{1- \tan^2 20}} tan ⁡ 65 − 2 tan ⁡ 40 = ( 1 + t a n 20 ) 2 − 4 tan ⁡ 20 1 − tan ⁡ 2 20 \tan65 - 2\tan40=\frac{(1+tan20)^2-4\tan20}{1-\tan^2 20} ( 1 − tan ⁡ 20 ) 2 ( 1 − t a n 20 ) ( 1 + t a n 20 ) = 1 − tan ⁡ 20 1 + tan ⁡ 20 = t a n ( 45 − 20 ) = t a n 25 \frac{(1-\tan20)^2}{(1-tan20)(1+tan20)}=\frac{1- \tan20}{1+\tan20}=tan(45-20)=tan25 Hence tan ⁡ − 1 ( t a n 25 ) = 25 \tan^{-1}(tan25)=\huge\color{#69047E}{25}

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= arctan ⁡ { 2.1445 − 2 ( 0.8390 ) } = arctan ⁡ { 2.1445 − 1.6781 } = arctan ⁡ { 0.4663 } = 25 =\arctan { \{ 2.1445-2(0.8390)\} } =\arctan { \{ 2.1445-1.6781\} } =\arctan { \{ 0.4663\} } =25

Akhil Garg - 5 years, 3 months ago
Amed Lolo
Feb 22, 2016

tan65=tan(90-25)=sin(90-25)÷cos(90-25). =cos 25÷sin25 ,,,,tan40=sin(90-50)÷ cos(90-50) .=cos50÷sin50=cos2(25)÷sin2(25). cos25\sin25-2cos2(25)\2cos25sin25. ={1÷sin25}×(cos25- cos2(25)\cos25). cos2(25)\cos25=2cos^2(25)-1\cos25=2cos25-1\cos25, {1\si25}×(cos25-2cos25+(1\cos25) =1\sin25×(1\cos25-cos25)={1\sin25}×{sin^2 (25)\cos25}=sin25÷cos25=tan25 so arctan(tan25)=25#####

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65 - 40 = 25

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Pulkit Gupta
Jan 8, 2016

Write

tan ⁡ 65 = tan ⁡ ( 45 + 20 ) = 1 + t a n 20 1 − t a n 20 \large \tan 65 = \tan (45+20) = \frac{1 + tan20}{1-tan20} ,

2 tan ⁡ 40 = 2 ( 2 tan ⁡ 20 ) 1 − tan ⁡ 2 20 \large 2 \tan 40 = \frac{2(2 \tan 20)}{1- \tan^2 20}

& simplify the expression to obtain ( 1 − tan ⁡ 20 ) 2 1 − tan ⁡ 2 20 \large \frac{(1-\tan 20)^2}{1- \tan^2 20} = ( 1 − tan ⁡ 20 ) 1 + tan ⁡ 20 \large \frac{(1-\tan 20)}{1+ \tan 20} = tan ⁡ ( 45 − 20 ) \large \tan (45-20) = tan ⁡ 25 \large \tan 25

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