In a triangle ABC we define x - tan B-C2 tan A2 y - tan C-A2 tan B2 and z - tan A-B2 tan C2 then the value of x-y-z in terms of x- y- z is

The correct option is D xyz tan B C2 = b cb+c cot A2 , #160;Napier's analogy ∴x1 = tan B C2 tan A2 = b cb+c #160; ∴ 1+x1 x #160;= 2b2c ...

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Byju's Answer

Standard XII

Mathematics

Napier’s Analogy

In a triangle...

Question

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

xyz

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- xyz

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2xyz

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none

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Solution

The correct option is D - xyz
$t a n \frac{B - C}{2} = \frac{b - c}{b + c} c o t \frac{A}{2}$ , Napier's analogy
$∴ \frac{x}{1} = t a n \frac{B - C}{2} t a n \frac{A}{2} = \frac{b - c}{b + c} ∴ \frac{1 + x}{1 - x} = \frac{2 b}{2 c} = \frac{b}{c}$
Similarly $\frac{1 + y}{1 - y} = \frac{c}{a}$ and $\frac{1 + z}{1 - z} = \frac{a}{b}$
$∴ \frac{(1 + x) (1 + y) (1 + z)}{(1 - x) (1 - y) (1 - z)} = \frac{b}{c} . \frac{c}{a} . \frac{a}{b} = 1$
or $1 + \sum x + \sum x y + x y z = 1 - \sum x + \sum x y - x y z$
$∴ 2 \sum x = - 2 x y z ∴ \sum x - x y z \Rightarrow (b)$

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Similar questions

Q. 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.

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. 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 = ?

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

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

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In a triangle ABC we definex=tanB−C2tanA2y=tanC−A2tanB2 and z=tanA−B2tanC2 then the value of x+y+z (in terms of x, y, z) is

The correct option is D - xyztanB−C2=b−cb+ccotA2, Napier's analogy∴x1=tanB−C2tanA2=b−cb+c∴1+x1−x=2b2c=bcSimilarly 1+y1−y=ca and 1+z1−z=ab∴(1+x)(1+y)(1+z)(1−x)(1−y)(1−z)=bc.ca.ab=1or 1+∑x+∑xy+xyz=1−∑x+∑xy−xyz∴2∑x=−2xyz∴∑x−xyz⇒(b)

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