If x+y-z+xyz=0, then 2x1-x2+2y1-y2-2z1-z2 is equal to

Let x=tanA,y=tanB,−z=tanC. Then (x+y−z)=−xyz⇒tanA+tanB+tanC=tanAtanBtanC⇒A+B+C=π ⇒2 A+2 B=2π−2C⇒tan(2A+2B)=tan(2π−2C)=−tan2C⇒tan2 A+tan2 B+tan2C=tan2 A⋅tan2 B⋅tan2C⇒2tanA1−tan2 A+2tanB1−tan2 B+2tanC1−tan2C=2tanA1−tan2 A⋅2tanB1−tan2 B⋅2tanC1−tan2C Put the value of tanA,tanB,tanC, we get ⇒2x1−x2+2y1−y2−2z1−z2=−8xyz1−x21−y21−z2

If x+y-z+xyz=0, then 2x1-x2+2y1-y2-2z1-z2 is equal to

Let x=tanA,y=tanB,−z=tanC. Then (x+y−z)=−xyz⇒tanA+tanB+tanC=tanAtanBtanC⇒A+B+C=π ⇒2 A+2 B=2π−2C⇒tan(2A+2B)=tan(2π−2C)=−tan2C⇒tan2 A+tan2 B+tan2C=tan2 A⋅tan2 B⋅tan2C⇒2tanA1−tan2 A+2tanB1−tan2 B+2tanC1−tan2C=2tanA1−tan2 A⋅2tanB1−tan2 B⋅2tanC1−tan2C Put the value of tanA,tanB,tanC, we get ⇒2x1−x2+2y1−y2−2z1−z2=−8xyz1−x21−y21−z2

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

$If x + y - z + x y z = 0, then$ $\frac{2 x}{1 - x^{2}} + \frac{2 y}{1 - y^{2}} - \frac{2 z}{1 - z^{2}} is equal$ to

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$\frac{x y z}{[(1 - x^{2}) (1 - y^{2}) (1 - z^{2})]}$

$\frac{- x y z}{[(1 - x^{2}) (1 - y^{2}) (1 - z^{2})]}$

$\frac{8 x y z}{[(1 - x^{2}) (1 - y^{2}) (1 - z^{2})]}$

$\frac{- 8 x y z}{[(1 - x^{2}) (1 - y^{2}) (1 - z^{2})]}$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} L e t x = \tan A, y = \tan B, - z = \tan C . Then (x + y - z) = - x y z \\ \Rightarrow \tan A + \tan B + \tan C = \tan A \tan B \tan C \\ \Rightarrow A + B + C = π \Rightarrow 2 A + 2 B = 2 π - 2 C \\ \Rightarrow \tan (2 A + 2 B) = \tan (2 π - 2 C) = - \tan 2 C \\ \Rightarrow \tan 2 A + \tan 2 B + \tan 2 C = \tan 2 A \cdot \tan 2 B \cdot \tan 2 C \\ \Rightarrow \frac{2 \tan A}{1 - \tan^{2} A} + \frac{2 \tan B}{1 - \tan^{2} B} + \frac{2 \tan C}{1 - \tan^{2} C} = \frac{2 \tan A}{1 - \tan^{2} A} \cdot \frac{2 \tan B}{1 - \tan^{2} B} \cdot \frac{2 \tan C}{1 - \tan^{2} C} \\ Put the value of \tan A, \tan B, \tan C, we get \\ \Rightarrow \frac{2 x}{1 - x^{2}} + \frac{2 y}{1 - y^{2}} - \frac{2 z}{1 - z^{2}} = - \frac{8 x y z}{(1 - x^{2}) (1 - y^{2}) (1 - z^{2})} \end{array}$