Tan35°+tan10°tan35°+tan10°=?

2 Answers
P dilip_k
Mar 21, 2017

We know

#tan45^@=1#

#=>tan(35^@+10^@)=1#

#=>(tan35^@+tan10^@)/(1-tan35^@tan10^@)=1#

#=>tan35^@+tan10^@=1-tan35^@tan10^@#

#=>tan35^@+tan35^@tan10^@+tan10^@=1#

Ratnaker Mehta
Mar 21, 2017

#1.#

Explanation:

Knowing that, #tan(A+B)=(tanA+tanB)/(1-tanAtanB),# we plug in,

#A=35^@, and, B=10^@# to get,

#tan(35^@+10^@)=(tan35^@+tan10^@)/(1-tan35^@tan10^@).#

#because, tan (35^@+10^@)=tan45^@=1,#

#:. 1=(tan35^@+tan10^@)/(1-tan35^@tan10^@).#

#:. 1-tan35^@tan10^@=tan35^@+tan10^@#

#rArr 1=tan35^@tan10^@+tan35^@+tan10^@#

Enjoy Maths.!

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