If the sides a-b-c of any triangle be in G-P- then prove that x-y-z are also in G-P- where x- b 2 - c 2 tan B- tan C tan B-tan C- y- c 2 - a 2 tan C- tan A tan C-tan Aand z- a 2 - b 2 tan A -tan B tan A-tan B-

x=( b ^ 2 c ^ 2 ) sin( B+C ) sin( B C ) x=( b ^ 2 c ^ 2 ) sin ^ 2 ( B+C ) sin ^ 2 B sin ^ 2 C . k^ 2 ( sin ^ 2 B sin ^ 2 C ) Similarly ...

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Standard XII

Mathematics

Integration as Antiderivative

If the sides ...

Question

If the sides $a, b, c$ of any triangle be in $G . P .$ , then prove that $x, y, z$ are also in $G . P .$ where
$x = (b^{2} - c^{2}) \frac{tan B + tan C}{tan B - tan C}$ ,
$y = (c^{2} - a^{2}) \frac{tan C + tan A}{tan C - tan A}$
and $z = (a^{2} - b^{2}) \frac{tan A + tan B}{tan A - tan B}$ .

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Solution

$x = (b^{2} - c^{2}) \frac{sin (B + C)}{sin (B - C)}$
$x = (b^{2} - c^{2}) \frac{{sin}^{2} (B + C)}{{sin}^{2} B - {sin}^{2} C} . k^{2} ({sin}^{2} B - {sin}^{2} C)$
Similarly $y = b^{2}$ and $z = c^{2}$
Obviously $a^{2}, b^{2}, c^{2}$ are also in $G . P$ as $a, b, c$ are in $G . P$ .

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If the sides a,b,c of any triangle be in G.P., then prove that x,y,z are also in G.P. wherex=(b2−c2)tan B+tan Ctan B−tan C,y=(c2−a2)tan C+tan Atan C−tan Aand z=(a2−b2)tan A+tan Btan A−tan B.

x=(b2−c2)sin(B+C)sin(B−C)x=(b2−c2)sin2(B+C)sin2B−sin2C.k2(sin2B−sin2C)Similarly y=b2 and z=c2Obviously a2,b2,c2 are also in G.P as a,b,c are in G.P.

If the sides $a, b, c$ of any triangle be in $G . P .$ , then prove that $x, y, z$ are also in $G . P .$ where
$x = (b^{2} - c^{2}) \frac{tan B + tan C}{tan B - tan C}$ ,
$y = (c^{2} - a^{2}) \frac{tan C + tan A}{tan C - tan A}$
and $z = (a^{2} - b^{2}) \frac{tan A + tan B}{tan A - tan B}$ .

$x = (b^{2} - c^{2}) \frac{sin (B + C)}{sin (B - C)}$
$x = (b^{2} - c^{2}) \frac{{sin}^{2} (B + C)}{{sin}^{2} B - {sin}^{2} C} . k^{2} ({sin}^{2} B - {sin}^{2} C)$
Similarly $y = b^{2}$ and $z = c^{2}$
Obviously $a^{2}, b^{2}, c^{2}$ are also in $G . P$ as $a, b, c$ are in $G . P$ .