Question

Tan35°+tan10°tan35°+tan10°=?

Answer 1

We know

`tan45^@=1`

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

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

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

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

Answer 2

`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.!