The sides of a triangle are such that a1-m2n2-bm2-n2-c1-m21-n2Prove that A- 2 tan -1mn- B-2 tan-1mn and -mnbcm2-n2

From the given ratios, we have a+b(a+m^2)(1+n^2)=a b(1 m^2)(1 n^2)=c(1 m^2)(1+n^2) a+bc=1+m^21 m^2, a bc=a n^21+n^2 Now by sine rule we get cosA B2 ...

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Standard XII

Mathematics

Property 2

The sides of ...

Question

The sides of a triangle are such that $\frac{a}{1 + m^{2} n^{2}} = \frac{b}{m^{2} + n^{2}} = \frac{c}{(1 - m^{2}) (1 + n^{2})}$
Prove that $A = 2 t a n^{- 1} \frac{m}{n}, B = 2 t a n^{- 1} (m n)$ and $△ = \frac{m n b c}{m^{2} + n^{2}}$

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Solution

From the given ratios, we have
$\frac{a + b}{(a + m^{2}) (1 + n^{2})} = \frac{a - b}{(1 - m^{2}) (1 - n^{2})} = \frac{c}{(1 - m^{2}) (1 + n^{2})}$
$\frac{a + b}{c} = \frac{1 + m^{2}}{1 - m^{2}}, \frac{a - b}{c} = \frac{a - n^{2}}{1 + n^{2}}$
Now by sine rule we get
$\frac{c o s \frac{A - B}{2}}{c o s \frac{A + B}{2}} = \frac{1 + m^{2}}{1 - m^{2}}, \frac{s i n \frac{A - B}{2}}{s i n \frac{A + B}{2}} = \frac{1 - n^{2}}{1 + n^{2}}$
Apply Compo. & Divi. on both.
$t a n \frac{A}{2} t a n \frac{B}{2} = m^{2}, c o t \frac{A}{2} t a n \frac{B}{2} = n^{2}$
Multiplying and dividing them, we get
$∴ t a n^{2} \frac{A}{2} = \frac{m^{2}}{n^{2}}, t a n^{2} \frac{B}{2} = m^{2} n^{2}$
$∴ A = 2 t a n^{- 1} \frac{m}{n}, B = 2 t a n^{- 1} (m n)$
$△ = \frac{1}{2} b c s i n A = \frac{1}{2} b c . \frac{2 t a n (A / 2)}{1 + t a n^{2} (A / 2)}$ etc.

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The sides of a triangle are such thata1+m2n2=bm2+n2=c(1−m2)(1+n2)Prove that A=2tan−1mn,B=2tan−1(mn) and △=mnbcm2+n2

The sides of a triangle are such that $\frac{a}{1 + m^{2} n^{2}} = \frac{b}{m^{2} + n^{2}} = \frac{c}{(1 - m^{2}) (1 + n^{2})}$
Prove that $A = 2 t a n^{- 1} \frac{m}{n}, B = 2 t a n^{- 1} (m n)$ and $△ = \frac{m n b c}{m^{2} + n^{2}}$