# Painless Trigonometry: (December 14 – December 21, 2013)

**Sum of tangents**

*Given:*

Find the value of

*Comment your answer below.*

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1. Arsh Singh. yourmathguru.com

2. Muthu Krishna.Kle college , Bengaluru

3. Jeffry Robles, University of the Philippines – Diliman

4. K.Rahul Mohideen

5. Spandan B, Pace Jr. Science College, India

6. Sebastian Pimber. Colombia

7. Kennard Ong Sychingping. De La Salle University

8.Jhay Dela Cruz. PUP-Taguig

9. Kenny Wong, Jurong Junior College, Singapore

10. Erick Ocampo- Manila Science High School

11. Jia Syuen- SMJK Sam Tet, Malaysia

### Dan

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20

tan(x)+tan(y)=4

1/tan(x)+1/tan(y)=5 => tan(x).tan(y)=(tan(x)+tan(y))/5 =>tan(x).tan(y)=4/5

tan(x+y)=(tan(x)+tan(y))/(1-tan(x).tan(y))

tan(x+y)=4/(1-4/5)

tan(x+y)=20

Spandan B, Pace Jr. Science College, India-

tan(x) +tan(y)/ tan(x)tan(y)=5

Thus, tan(x)tan(y)=4/5…

tan(x+y)=20

20

Pup-taguig

-1/2

tan(x + y) = (tan(x) + tan(y))/(1 – tan(x)tan(y))

= 4/(1 – (tan(y) + tan(x))/(cot(x) + cot(y)))

= 4/(1 – 4/5)

= 20.

1/ tan x + 1/tan y = 5

(tan x + tan y) / ( tan x)(tan y)= 5

4 / (tan x)(tan y) = 5

(tan x)(tan y) = 4/5

tan (x+y ) = (tan x + tan y) / [ 1 – (tan x)(tan y)

tan (x+y) = 4 / [ 1 – 4/5]

tan(x+y) = 4 / (1/5) = 20

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