tan (45+x)+ tan (45-x) =1/2sinxcosx
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tan (45+x)- tan (45-x)
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Use tan(a+b) = (tana+tanb)/(1-tanatanb)tan(a+b)=tana+tanb1−tanatanb
and tan(a-b) = (tana-tanb)/(1+tanatanb)tan(a−b)=tana−tanb1+tanatanb
and tan (2a) = tan(a+a) = (2tana)/(1-tan^2a)tan(2a)=tan(a+a)=2tana1−tan2a
And also use: tan (pi/4) =1tan(π4)=1.
Other than that it's just algebra.
Once you get to
(tanx + 1)/(1-tanx)-(1-tanx)/(1+tanx)tanx+11−tanx−1−tanx1+tanx
you'll want a common denominator, so you get
((tanx+1)^2-(1-tanx)^2)/(1-tan^2x)(tanx+1)2−(1−tanx)21−tan2x
and tan(a-b) = (tana-tanb)/(1+tanatanb)tan(a−b)=tana−tanb1+tanatanb
and tan (2a) = tan(a+a) = (2tana)/(1-tan^2a)tan(2a)=tan(a+a)=2tana1−tan2a
And also use: tan (pi/4) =1tan(π4)=1.
Other than that it's just algebra.
Once you get to
(tanx + 1)/(1-tanx)-(1-tanx)/(1+tanx)tanx+11−tanx−1−tanx1+tanx
you'll want a common denominator, so you get
((tanx+1)^2-(1-tanx)^2)/(1-tan^2x)(tanx+1)2−(1−tanx)21−tan2x
Answered by
3
Use tan(a+b) = (tana+tanb)/(1-tanatanb)tan(a+b)=tana+tanb1−tanatanb
and tan(a-b) = (tana-tanb)/(1+tanatanb)tan(a−b)=tana−tanb1+tanatanb
and tan (2a) = tan(a+a) = (2tana)/(1-tan^2a)tan(2a)=tan(a+a)=2tana1−tan2a
And also use: tan (pi/4) =1tan(π4)=1.
Other than that it's just algebra.
Once you get to
(tanx + 1)/(1-tanx)-(1-tanx)/(1+tanx)tanx+11−tanx−1−tanx1+tanx
you'll want a common denominator, so you get
((tanx+1)^2-(1-tanx)^2)/(1-tan^2x)(tanx+1)2−(1−tanx)21−tan2x
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