Question

Which one is correct for Napier’s Analogy?

(a) tan (A/2) = (b – c)/(b + c) (tan(B + C)/2)

(b) tan (A/2) = (b – c)/(b + c) (cot(B – C)/2)

(c) tan (A/2) = (b + c)/(b – c) (cot(B – C)/2)

(d) tan (A) = (b – c)/(b + c) (tan(B – C)/2)

I have been asked this question in a national level competition.

This interesting question is from Trigonometric Equations topic in division Trigonometric Functions of Mathematics – Class 11

Answer 1

Correct answer is (b) tan (A/2) = (b – c)/(b + c) (cot(B – C)/2)

For explanation I would say: According to the law of sines, in any triangle ABC,

a/sinA = b/sinB = c/sinC

So, a/b = sinA/sinB

(a + b)/(a – b) = (sinA + sinB)/( sinA – sinB)

=> (a + b)/(a – b) = (2 sin((a + b)/2) cos((a – b)/2))/ (2 sin((a + b)/2) sin((a – b)/2))

=> (a + b)/(a – b) = (tan(A + B)/2)/(tan(A – B)/2)

=> (tan(A + B)/2) = (a + b)/(a – b) (tan(A – B)/2)

=> (tan(π/2 + C/2)) = (a + b)/(a – b) (tan(A – B)/2)

=> cot (C/2) = (a + b)/(a – b) (tan(A – B)/2)

=> tan (C/2) = (a – b)/(a + b) (tan(A – B)/2)

Similarly, tan (A/2) = (b – c)/(b + c) (cot(B – C)/2).