Question

In ΔABC, prove that $\frac{a^{2}\ +\ b^{2}\ -\ c^{2}}{c^{2}\ +\ a^{2}\ -\ b^{2}} = \frac{tan\ B}{tan\ C}$.

Answer 1

Solution :

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We know that ,

i ) SinB = b/2R

ii ) cosB = ( a²+c²-b²)/2ac

iii ) abc/R = 4∆

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1 ) tan B

= sinB/cosB

= ( b/2R )/[ ( a²+c²-b² )/2ac ]

= (b/2R)×[ (2ac )/(a²+c²-b²)]

= ( abc/R )/( a²+c²-b² )

= ( 4∆ )/( a²+c²-b² )

Similarly ,

tanC = ( 4∆ )/( a²+b²-c² )

Now ,

RHS = tanB/tanC

= [(4∆)/( a²+c²-b² )]/[(4∆)/(a²+b²-c²)]

= ( a²+b²-c² )/( a²+c²-b² )

= LHS

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Answer 2

HELLO DEAR,

Answer:

Step-by-step explanation:

We know:- SinB = b/2R , cosB = ( a²+c²-b²)/2ac and abc/R = 4∆

now,

tan B

= sinB/cosB

= ( b/2R )/[ ( a²+c²-b² )/2ac ]

= (b/2R)×[ (2ac )/(a²+c²-b²)]

= ( abc/R )/( a²+c²-b² )

= ( 4∆ )/( a²+c²-b² )----------( 1 )

Similarly ,

tanC = ( 4∆ )/( a²+b²-c² )---------( 2 )

from---------( 1 ) & ----------( 2 )

tanB/tanC = [(4∆)/( a²+c²-b² )]/[(4∆)/(a²+b²-c²)]

= ( a²+b²-c² )/( a²+c²-b² )

I HOPE IT'S HELP YOU DEAR,

THANKS