If \; x/2=\; y/3=\; z/5 and x+y+z=π, then the value of ² x+² y+² z is:

Correct Answer is : 38/3{38}/338/3

Correct Answer is : 38/3{38}/338/3

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NIMCET 2018 Question Paper with solutions

Question : 5 of 120

Marks: +1, -0

\;\frac{\tan x}{2}=\;\frac{\tan y}{3}=\;\frac{\tan z}{5}

x+y+z=\pi

\tan^2 x+\tan^2 y+\tan^2 z

is:

Solution:

Concept:

Trigonometric Identities:

\tan(A+B)=\;\frac{\tan A+\tan B}{1-\tan A\tan B}

\tan(n\pi+\theta)=\tan\theta

\tan(-\theta)=-\tan\theta

Calculations:

It is given that

\;\frac{\tan x}{2}=\;\frac{\tan y}{3}=\;\frac{\tan z}{5}=k

(say).

\therefore\tan x=2k,\tan y=3k

and

\tan z=5k

It is also given that

x+y+z=\pi

\Rightarrow x+y=\pi-z

\Rightarrow\tan(x+y)=\tan[\pi+(-z)]

\Rightarrow\;\frac{\tan x+\tan y}{1-\tan x\tan x}=\tan(-z)=-\tan z

\Rightarrow\tan x+\tan y=-\tan z+\tan x\tan y\tan z

\Rightarrow\tan x+\tan y+\tan z=\tan x\tan y\tan z

Substituting the values in terms of

k

from the above result, we get:

\;\;2k+3k+5k=(2k)(3k)(5k)

\Rightarrow10k=30k^3

\Rightarrow k^2=\;\frac{1}{3}

Now,

\tan^2 x+\tan^2 y+\tan^2 z

\;\;=(2k)^2+(3k)^2+(5k)^2

\;\;=4k^2+9k^2+25k^2

\;\;=\;\frac{38}{3}.

\frac{38}{3}

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