In ΔABC, m∠C=90 and m∠A=m∠B,Is tanA=tanB?
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In case that C is obtuse,
Then C>90.
So A+B msut be less than 90... (because in a triangle A+B+C=180, and C is above 90)
So tan(A+B) must be postivie... (Since A+B is acute < 90)
tan(A+B) = (tanA+tanB)/(1-tanAtanB)
So 1-tanAtanB>0..... (Only then tan(A+B) can be positive)
Hence tanAtanB<1
Then C>90.
So A+B msut be less than 90... (because in a triangle A+B+C=180, and C is above 90)
So tan(A+B) must be postivie... (Since A+B is acute < 90)
tan(A+B) = (tanA+tanB)/(1-tanAtanB)
So 1-tanAtanB>0..... (Only then tan(A+B) can be positive)
Hence tanAtanB<1
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In ∆ABC ,
m<C = 90°
m<A = m<B = x ,
m<A + m<B + m<C = 180°
[ Angle sum property ]
=> x + x + 90° = 180°
=> 2x = 180° - 90°
=> 2x = 90°
=> x = 90°/2 = 45°
Therefore ,
m<A = m<B = 45°
Now ,
tan A = tan 45° = 1 ----( 1 )
tan B = tan 45° = 1 ----( 2 )
From ( 1 )& ( 2 ), we get
tan A = tan B
••••••
m<C = 90°
m<A = m<B = x ,
m<A + m<B + m<C = 180°
[ Angle sum property ]
=> x + x + 90° = 180°
=> 2x = 180° - 90°
=> 2x = 90°
=> x = 90°/2 = 45°
Therefore ,
m<A = m<B = 45°
Now ,
tan A = tan 45° = 1 ----( 1 )
tan B = tan 45° = 1 ----( 2 )
From ( 1 )& ( 2 ), we get
tan A = tan B
••••••
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