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Which one is correct for Napier’s Analogy?(a) tan (B/2) = (c – a)/(c + a) (cot(C + A)/2)(b) tan (B/2) = (c – a)/(c + a) (cot(C – A)/2)(c) tan (B/2) = (c + a)/(c – a) (cot(C – A)/2)(d) tan (B) = (c – a)/(c + a) (cot(C – A)/2)I have been asked this question in an internship interview.Asked question is from Trigonometric Equations in portion Trigonometric Functions of Mathematics – Class 11

Answer»

The correct ANSWER is (b) tan (B/2) = (c – a)/(c + a) (cot(C – A)/2)

EASIEST explanation: 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) (cot(A – B)/2)

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


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