How do I get from \(\displaystyle{\frac{{\sqrt{{{3}}}}}{{{2}}}}+{\frac{{{1}}}{{{2}}}}{\tan{{x}}}-{2}{\tan{{x}}}\cdot{\frac{{\sqrt{{{3}}}}}{{{2}}}}\) to \(\displaystyle{\frac{{\sqrt{{{3}}}}}{{{2}\sqrt{{{3}}}-{1}}}}={\tan{{x}}}\) and then

Kiara Haas

Kiara Haas

Answered question

2022-03-25

How do I get from
32+12tanx2tanx32
to
3231=tanx
and then to
11tanx=6+3

Answer & Explanation

pastuh7vka

pastuh7vka

Beginner2022-03-26Added 13 answers

I will try to make it a bit more elaborate since you are going to need to be able to manipulate equations like this all the time.
Multiply both sides by 2:
3+tan(x)=2tan(x)3
Subtract tan(x) from both sides in order isolate tan(x)
3=2tan(x)3tan(x)
You have two terms with tan(x) on the RHS so you can factor out tan(x)
3=tan(x)(231)
Divide both sides by (231)
Now you can multiply the fraction on the RHS by 23+123+1 in order to get rid of the sqare root in the denominator. Remember you are just multiplying by 1 so it doesn't change the equation
tan(x)=323123+123+1
Just do a bit of multiplication and you will get
tan(x)=6+31111tan(x)=6+3
Ishaan Stout

Ishaan Stout

Beginner2022-03-27Added 14 answers

Simple: isolate tanx in one side. You get:
2312tanx=32 whence tanx=3231
Then rationalise the denominator:
tanx=3(23+112-1=6+311

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