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 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

- xyz

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

2xyz

No worries! We‘ve got your back. Try BYJU‘S free classes today!

none

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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 $1 + \sum x + \sum x y + x y z = 1 - \sum x + \sum x y - x y z$
$∴ 2 \sum x = - 2 x y z ∴ \sum x - x y z \Rightarrow (b)$

Suggest Corrections

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

Join BYJU'S Learning Program

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