Question #a7c99

2 Answers
turksvids
Jan 3, 2018

see below.

Explanation:

The difference formula for tangent is:

#tan(x-y) = (tanx-tany)/(1+tanxtany)#

So,
#tan(15^circ) = tan(45^circ-30^circ) #

#=(tan(45^circ)-tan(30^circ))/(1+tan(45^circ)tan(30^circ))#

We know that #tan(45^circ)=1# and #tan(30^circ)=sqrt(3)/3#, so substituting those values:

#=(1-(sqrt(3)/3))/(1+(1)(sqrt(3)/3))#

Getting common denominators within the numerator and denominator:

#=((3-sqrt(3))/3)/((3+sqrt(3))/3)#

Simplifying complex fractions:

#=(3-sqrt(3))/(3+sqrt(3))#

Rationalizing the denominator (multiply by the conjugate of the denominator):

#=(3-sqrt(3))/(3+sqrt(3))*(3-sqrt(3))/(3-sqrt(3)) #

Simplifying numerator and denominator:

#(9-6sqrt(3)+3)/(9-3)#

Collecting like terms:

#(12-6sqrt(3))/(6)#

Simplifying:

#2-sqrt(3)#

Jim G.
Jan 3, 2018

#"see explanation"#

Explanation:

#"using the "color(blue)"difference formula for tan"#

#•color(white)(x)tan(A-B)=(tanA-tanB)/(1-tanAtanB)#

#"we can express "15^@" as "45^@-30^@#

#rArrtan15^@=tan(45-30)^@#

#rArrtan(45-30)#

#=(tan45-tan30)/(1+tan45tan30)#

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

#"multiply numerator/denominator by "sqrt3#

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

#"multiply numerator/denominator by the "color(blue)"conjugate"#
#"of the denominator"#

#"the "color(blue)"conjugate ""of "sqrt3+1" is "sqrt3color(red) (-)1#

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

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

#=(4-2sqrt3)/2=cancel(4)^2/cancel(2)^1-cancel(2)/cancel(2)sqrt3=2-sqrt3#

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