Question

What is `{sin10°+sin20°}/{cos10°+cos20°} =` ?

Answer 1

`(sin20^@+sin10^@)/(cos20^@+cos10^@)=2-sqrt3`

Explanation:

As `sinA+sinB=2sin((A+B)/2)cos((A-B)/2)`

and `cosA+cosB=2cos((A+B)/2)cos((A-B)/2)`

Hence `(sin10^@+sin20^@)/(cos10^@+cos20^@)`

= `(sin20^@+sin10^@)/(cos20^@+cos10^@)`

= `(2sin((20^@+10^@)/2)cos((20^@-10^@)/2))/(2cos((20^@+10^@)/2)cos((20^@-10^@)/2))`

= `(sin15^@cos5^@)/(cos15^@cos5^@)`

= `tan15^@=tan(45^@-30^@)`

= `(tan45^@-tan30^@)/(1+tan45^@tan30^@)`

= `(1-1/sqrt3)/(1+1/sqrt3)`

= `(sqrt3-1)/(sqrt3+1)xx(sqrt3-1)/(sqrt3-1)`

= `(3+1-2sqrt3)/(3-1)`

= `2-sqrt3`

Answer 2

`{sin10°+sin20°}/{cos10°+cos20°}`

#={2sin15°cos5°}/{2cos15°cos5°} #

`=tan15^@`

`=tan(45^@-30^@)`

`=(tan45^@-tan30^@)/(1+tan45^@tan30^@)`

`=(1-1/sqrt3)/(1+1/sqrt3)`

`=(sqrt3-1)/(sqrt3+1)`

`=(sqrt3-1)^2/((sqrt3)^2-1)`

`=(4-2sqrt3)/2=2-sqrt3`