If in a triangle ABC- we define x - tan B-C-2 tanA-2- y-tanC-A-2tanB-2- z- tanA-B-2tanC-2then show that x-y-z - -xyz-

tan B C2= b cb+ccotA2 ∴x1tanB C2tanA/2=b c/b+c ∴1+x1 x=2b2c . similarly 1+y1 x=2c2a, 1+z1 z=2a2b ∴(1+x)(1+y)(1+z)(1 x)(1 y)(1 z)=1 or ...

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Byju's Answer

Standard XII

Mathematics

General Solution of Trigonometric Equation

If in a trian...

Question

If in a triangle ABC, we define
$x = t a n \frac{B - C}{2} t a n \frac{A}{2}$ , $y = t a n \frac{C - A}{2} t a n \frac{B}{2}$ ,
$z = t a n \frac{A - B}{2} t a n \frac{C}{2}$
then show that x+y+z = -xyz.

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Solution

$t a n \frac{B - C}{2} = \frac{b - c}{b + c} c o t \frac{A}{2}$

$∴ \frac{x}{1} t a n \frac{B - C}{2} t a n \frac{A}{2} = \frac{b - c}{b + c}$

$∴ \frac{1 + x}{1 - x} = \frac{2 b}{2 c}$ . similarly

$\frac{1 + y}{1 - x} = \frac{2 c}{2 a}, \frac{1 + z}{1 - z} = \frac{2 a}{2 b}$

$∴ \frac{(1 + x) (1 + y) (1 + z)}{(1 - x) (1 - y) (1 - z)} = 1$

or $1 + \sum x + \sum x y + x y z = 1 - \sum x + \sum x y - x y z$

or $2 \sum x = - 2 x y z$ or $\sum x = - x y z$

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Similar questions

Q. In a triangle ABC we define

x = t a n \frac{B - C}{2} t a n \frac{A}{2}

y = t a n \frac{C - A}{2} t a n \frac{B}{2}

and

z = t a n \frac{A - B}{2} t a n \frac{C}{2}

then the value of

x + y + z

(in terms of x, y, z) is

Q. If

x = tan (\frac{B - C}{2}) tan \frac{A}{2}, y = tan (\frac{C - A}{2}) tan \frac{B}{2}, z = tan (\frac{A - B}{2}) tan \frac{C}{2}

then

x + y + z = ?

In a triangle $A B C$ we define
$x = t a n \frac{B - C}{2} t a n \frac{A}{2}, y = t a n \frac{C - A}{2} t a n \frac{B}{2}$ and $z = t a n \frac{A B}{2} t a n \frac{C}{2}$
Then the value of $x + y + z$ (in terms of $x, y, z$ ) is

Q. In

△ A B C

, let

x = tan \frac{B - C}{2} tan \frac{A}{2}, y = tan \frac{C - A}{2} tan \frac{B}{2}, z = tan \frac{A - B}{2} tan \frac{C}{2}

Then

x + y + z = K (x y z)

, where the value of

K

Q. In

Δ A B C

. If

x = tan (\frac{B - C}{2}) tan \frac{A}{2}, y = tan (\frac{C - A}{2}) tan \frac{B}{2}, z = tan (\frac{A - B}{2}) tan \frac{C}{2}

, then

x + y + z

(in terms of

x, y, z

only) is

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If in a triangle ABC, we definex=tanB−C2tanA2, y=tanC−A2tanB2,z=tanA−B2tanC2then show that x+y+z = -xyz.

tanB−C2=b−cb+ccotA2∴x1tanB−C2tanA2=b−cb+c∴1+x1−x=2b2c. similarly1+y1−x=2c2a,1+z1−z=2a2b∴(1+x)(1+y)(1+z)(1−x)(1−y)(1−z)=1or 1+∑x+∑xy+xyz=1−∑x+∑xy−xyzor 2∑x=−2xyz or ∑x=−xyz

If in a triangle ABC, we define
$x = t a n \frac{B - C}{2} t a n \frac{A}{2}$ , $y = t a n \frac{C - A}{2} t a n \frac{B}{2}$ ,
$z = t a n \frac{A - B}{2} t a n \frac{C}{2}$
then show that x+y+z = -xyz.