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Trigonometry

Question
Find
tan-1(1⁄2 tan 2A) + tan-1(cot A) + tan-1(cot3 A), when 0 < A < π/2

Explanation
- Analyze the first term
tan-1x varies from 0 to π/2 when x is positive
and varies from -π/2 to 0 when x is negative

tan x is positive when x is between 0 and π/2 and -π and -π/2
and is negative when x is between π/2 and π and 0 and -π/2

So the first term is positive from 0 to π/4 and negative from π/4 to π/2

Second and Third terms are always positive in the range of A.

Use the identity tan-1x + tan-1y = tan-1 (x+y)⁄(1-xy)
to add first and third terms

tan-1(1⁄2 tan 2A) + tan-1(cot3 A)
= tan-1 ( 2tanA/2(1-tan2A) + 1/tan3A ⁄ 1- 2tanA/2(1-tan2A) .1/tan3A)

Multiplying both numerator and denominator of the inside fraction by
(1-tan2A)tan3A

= tan-1 ( tanA.tan3A+ 1 - tan2A ⁄ (1-tan2A)tan3A-tan A)

= tan-1 ( tan4A+ 1 - tan2A ⁄ tan3A-tan5A-tan A)

= tan-1 ( -1 ⁄ tanA)
= tan-1 (-cot A) , positive for 0 < A < π/4 and negative for π/4 < A < π/2

Now add the second term to the above result and use the identity used before
The Question reduces to tan-1 (-cot A)  +  tan-1(cot A)
= tan-1 ( -cot A + cot A ⁄ denominator)
= tan-1 0
This can either be 0 or π
As the first term is negative for π/4 < A < π/2
Sum of 3 terms = 0 for π/4 < A < π/2 and π for 0 < A < π/4
Variation of values of terms with angle A


Other problems of my interest can be found here
Exponential Remainder problem
Circles and Tangents

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