Vietnam
In the examination hall, Barbara tried hard to simplify an expression in cotangents. But she got confused and was not able to complete it properly. After the examination, she seeks the help of Michelle for simplifying it. Then what would be Barbara's findings, if the expression was:
@^ \dfrac{ 3cot^3x - cotx }{ cot^4x - 3cot^2x } = ? @^
Answer:
^@ tan3x ^@
- The expression can be simplified by first taking common terms out. Then the terms in the whole expression can be expressed in terms of the tangent to get the most simplified expression. After obtaining each term in the expression in tangent, ^@tan3x^@ from the given options can be expanded to obtain the result.
- The following expression can be simplified as: @^ \dfrac{ 3cot^3x - cotx }{ cot^4x - 3cot^2x } = \dfrac{ cotx(3cot^2x-1) }{ cot^2x(cot^2x - 3) } = \dfrac{ 3cot^2x - 1 }{ cotx(cot^2x - 3) } @^
- Now putting ^@ cotx = \dfrac{ 1 }{ tanx } ^@ and rewriting the expression in terms of tangent. @^ \begin{align} & \dfrac{ 3cot^2x - 1 }{ cotx(cot^2x - 3) } \\ =& \dfrac{ tanx(3-tan^2x) }{ 1 - 3tan^2x } \\ =& \dfrac{ 3tanx - tan^3x }{ 1 - 3tan^2x } \\ =& \dfrac { 2tanx + tanx - tan^3x }{ 1 - tan^2x - 2tan^2x } \\ =& \dfrac { 2tanx + tanx(1 - tan^2x) }{ 1 - tan^2x - 2tan^2x } && \left[ \text{ Dividing whole equation by } 1 - tan^2x \right] \\ =& \dfrac { \dfrac { 2tanx }{ 1-tan^2x } + tanx }{ 1 - \dfrac{ 2tan^2x }{ 1-tan^2x } } && \left[ \text{ Using identity } tan2A = \dfrac{ 2tanA } { 1 - tan^2A } \right] \\ =& \dfrac{ tan2x + tanx }{ 1 - tan2x \times tanx } && \left[ \text{ Using identity } tan(A + B) = \dfrac{ tanA + tanB } { 1 - tanAtanB } \right] \\ =& tan(2x + x) \\ =& tan3x \end{align} @^
