If x, y and z are real numbers that satisfy the three equations \ array l (x)+ (y)+ (z)=6-( (x)+ (y)+ (z)) ^2(x)+ ^2(y)+ ^2(z)=6-( ^2(x)+ ^2(y)+ ^2(z)) ^3(x)+ ^3(y)+ ^3(z)=6-( ^3(x)+ ^3(y)+ ^3(z)) array . Find the value

Correct answer: 6 Let x=a, y=b, z=c Given system of equation is equal to aligned a+b+c & =6-(1)/(a)-(1)/(b)-(1)/(c) a^2+b^2+c^2 & =6-(1)/(a^2)-(1)/(b^2)-(1)/(c^2) a^3+b^3+c^3 & =6-(1)/(a^3)-(1)/(b^3)-(1)/(c^3) aligned From second equation we complete squares to get (a-(1)/(a))^2+(b-(1)/(b))^2+(c-(1)/(c))^2=0 a=b=c= ± 1 Now rearranging 3^ rd equation and adding 3 time first equation, we get aligned a^3+b^3+c^3+(1)/(a^2)+(1)/(b^2)+(1)/(c^2)+3(a+(1)/(a)+b+(1)/(b)+c+(1)/(c)) & =6+18 (a+(1)/(a))^3+(b+(1)/(b))^3+(c+(1)/(c))^3 & =24 aligned Therefore a=b=c=1 Hence, [( (x))/( (y))+( (y))/( (z))+( (z))/( (x))+3 (x) (y) (z)]=6

JEE Advanced

Exam: JEE Advanced
Subject: Mathematics
Chapter: Trigonometry

JEE Advanced · Mathematics · Trigonometry

If x, y and z are real numbers that satisfy the three equations \...

x, y

and

z

are real numbers that satisfy the three equations

\left\{

\begin{array}{l} \tan (x)+\tan (y)+\tan (z)=6-(\cot (x)+\cot (y)+\cot (z)) \\ \tan ^2(x)+\tan ^2(y)+\tan ^2(z)=6-\left(\cot ^2(x)+\cot ^2(y)+\cot ^2(z)\right) \\ \tan ^3(x)+\tan ^3(y)+\tan ^3(z)=6-\left(\cot ^3(x)+\cot ^3(y)+\cot ^3(z)\right) \end{array}

\right.

Find the value of the expression

\left(\frac{\tan (x)}{\tan (y)}+\frac{\tan (y)}{\tan (z)}+\frac{\tan (z)}{\tan (x)}+3 \tan (x) \tan (y) \tan (z)\right)

Correct answer

Step-by-step solution

Let

\tan x=a, \tan y=b, \tan z=c

Given system of equation is equal to \begin{aligned} a+b+c & =6-\frac{1}{a}-\frac{1}{b}-\frac{1}{c} \\ a^2+b^2+c^2 & =6-\frac{1}{a^2}-\frac{1}{b^2}-\frac{1}{c^2} \\ a^3+b^3+c^3 & =6-\frac{1}{a^3}-\frac{1}{b^3}-\frac{1}{c^3} \end{aligned} From second equation we complete squares to get

\left(a-\frac{1}{a}\right)^2+\left(b-\frac{1}{b}\right)^2+\left(c-\frac{1}{c}\right)^2=0 \Rightarrow a=b=c= \pm 1

Now rearranging

3^{\text {rd }}

equation and adding 3 time first equation, we get \begin{aligned} a^3+b^3+c^3+\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+3\left(a+\frac{1}{a}+b+\frac{1}{b}+c+\frac{1}{c}\right) & =6+18 \\ \left(a+\frac{1}{a}\right)^3+\left(b+\frac{1}{b}\right)^3+\left(c+\frac{1}{c}\right)^3 & =24 \end{aligned} Therefore

a=b=c=1

Hence, $\left[\frac{\tan (x)}{\tan (y)}+\frac{\tan (y)}{\tan (z)}+\frac{\tan (z)}{\tan (x)}+3 \tan (x) \tan (y) \tan (z)\right]=6

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