Question

in a triangle ABC b(by tangent rule)=​

Answer 1

Answer:

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Step-by-step explanation:

In any triangle ABC we have

⇒ bsinB = csinC

⇒ bc = sinBsinC

⇒ (b−cb+c) = sinB−sinCsinB+sinC, [Applying Dividendo and Componendo]

⇒ (b−cb+c) = 2cos(B+C2)sin(B−C2)2sin(B+C2)cos(B−C2)

⇒ (b−cb+c) = cot (B+C2) tan (B−C2)

⇒ (b−cb+c) = cot (π2 - A2) tan (B−C2), [Since, A + B + C = π ⇒ B+C2 = π2 - A2]

⇒ (b−cb+c) = tan A2 tan (B−C2)

⇒ (b−cb+c) =  tanB−C2cotA2

Therefore, tan (B−C2) = (b−cb+c) cot A2.                        Proved.

Similarly, we can prove that the formulae (ii) tan (C−A2)  = (c−ac+a) cot B2 and (iii) tan (A−B2) = (a−ba+b)  cot C2.

 

Alternative Proof law of tangents:

According to the law of sines, in any triangle ABC,                      

     asinA = bsinB = csinC

Let, asinA = bsinB = csinC = k

Therefore,

asinA = k, bsinB = k and csinC = k

⇒ a = k sin A, b = k sin B and c = k sin C ……………………………… (1)

Proof of formula (i) tan (B−C2) = (b−cb+c) cot A2

R.H.S. = (b−cb+c) cot A2

= ksinB−ksinCksinB+ksinC cot A2, [Using (1)]

= (sinB−sinCsinB+sinC) cot A2

= 2sin(B−C2)cos(B+c2)2sin(B+C2)cos(B−c2)

= tan (B−C2) cot (B+C2) cot A2

= tan (B−C2) cot (π2 - A2) cot A2, [Since, A + B + C = π ⇒ B+C2 = π2 - A2]

= tan (B−C2) tan A2 cot A2

= tan (B−C2) = L.H.S.

Similarly, formula (ii) and (iii) can be proved.