Question :- If f(x) = tanx, then prove that
[tex]\sf \: (i) \: \: f(x + y)= \: \dfrac{f(x) + f(y)}{1 - f(x) \: f(y)} \\ \\ [/tex]
[tex]\sf \: (ii) \: \: f(2x) = \: \dfrac{2 \: f(x)}{1 - {[ f(x)]}^{2} } \\ \\ \\ [/tex]
[tex]\large\underline{\sf{Solution-}}[/tex]
Given that, f(x) = tanx
Now, Consider
[tex]\sf \: f(x + y) \\ \\ [/tex]
[tex]\sf \: = \: tan(x + y) \\ \\ [/tex]
[tex]\sf \: = \: \dfrac{tanx + tany}{1 - tanx \: tany} \\ \\ [/tex]
[tex]\sf \: = \: \dfrac{f(x) + f(y)}{1 - f(x) \: f(y)} \\ \\ [/tex]
Hence,
[tex]\sf \: \sf\implies \boxed{ \bf{ \:f(x + y) = \: \dfrac{f(x) + f(y)}{1 - f(x) \: f(y)} \: }}\\ \\ [/tex]
[tex]\sf \: f(2x) \\ \\ [/tex]
[tex]\sf \: = \: tan2x \\ \\ [/tex]
[tex]\sf \: = \: \dfrac{2 \: tanx}{1 - {tan}^{2} x} \\ \\ [/tex]
[tex]\sf \: = \: \dfrac{2 \: tanx}{1 - {(tanx)}^{2}} \\ \\ [/tex]
[tex]\sf \: = \: \dfrac{2 \: f(x)}{1 - {[ f(x)]}^{2} } \\ \\ [/tex]
[tex]\sf \: \sf\implies \boxed{ \bf{ \:f(2x) = \: \dfrac{2 \: f(x)}{1 - {[ f(x)]}^{2} } \: }} \\ \\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
Additional Information
[tex]{ \boxed{ \begin{array}{c}\rm \: sin2x \: = 2 \: sinx \: cosx\:\\ & \rm \: cos2x = 1 - {2sin}^{2}x \\ & \rm \: cos2x = {2cos}^{2}x - 1 \\ & \rm \: cos2x = {cos}^{2}x - {sin}^{2}x \\ & \rm \:tan2x = \dfrac{2tanx}{1 - {tan}^{2} x} \\ & \rm \: sin2x = \dfrac{2tanx}{1 + {tan}^{2}x } \\ & \rm \:sin3x = 3sinx - {4sin}^{3}x \\ & \rm \: cos3x = {4cos}^{3}x - 3cosx \\ & \rm \: tan3x = \dfrac{3tanx - {tan}^{3} x}{1 - {3tan}^{2}x} \end{array}}} \\ \\[/tex]
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Answers & Comments
Question :- If f(x) = tanx, then prove that
[tex]\sf \: (i) \: \: f(x + y)= \: \dfrac{f(x) + f(y)}{1 - f(x) \: f(y)} \\ \\ [/tex]
[tex]\sf \: (ii) \: \: f(2x) = \: \dfrac{2 \: f(x)}{1 - {[ f(x)]}^{2} } \\ \\ \\ [/tex]
[tex]\large\underline{\sf{Solution-}}[/tex]
Given that, f(x) = tanx
Now, Consider
[tex]\sf \: f(x + y) \\ \\ [/tex]
[tex]\sf \: = \: tan(x + y) \\ \\ [/tex]
[tex]\sf \: = \: \dfrac{tanx + tany}{1 - tanx \: tany} \\ \\ [/tex]
[tex]\sf \: = \: \dfrac{f(x) + f(y)}{1 - f(x) \: f(y)} \\ \\ [/tex]
Hence,
[tex]\sf \: \sf\implies \boxed{ \bf{ \:f(x + y) = \: \dfrac{f(x) + f(y)}{1 - f(x) \: f(y)} \: }}\\ \\ [/tex]
Now, Consider
[tex]\sf \: f(2x) \\ \\ [/tex]
[tex]\sf \: = \: tan2x \\ \\ [/tex]
[tex]\sf \: = \: \dfrac{2 \: tanx}{1 - {tan}^{2} x} \\ \\ [/tex]
[tex]\sf \: = \: \dfrac{2 \: tanx}{1 - {(tanx)}^{2}} \\ \\ [/tex]
[tex]\sf \: = \: \dfrac{2 \: f(x)}{1 - {[ f(x)]}^{2} } \\ \\ [/tex]
Hence,
[tex]\sf \: \sf\implies \boxed{ \bf{ \:f(2x) = \: \dfrac{2 \: f(x)}{1 - {[ f(x)]}^{2} } \: }} \\ \\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
Additional Information
[tex]{ \boxed{ \begin{array}{c}\rm \: sin2x \: = 2 \: sinx \: cosx\:\\ & \rm \: cos2x = 1 - {2sin}^{2}x \\ & \rm \: cos2x = {2cos}^{2}x - 1 \\ & \rm \: cos2x = {cos}^{2}x - {sin}^{2}x \\ & \rm \:tan2x = \dfrac{2tanx}{1 - {tan}^{2} x} \\ & \rm \: sin2x = \dfrac{2tanx}{1 + {tan}^{2}x } \\ & \rm \:sin3x = 3sinx - {4sin}^{3}x \\ & \rm \: cos3x = {4cos}^{3}x - 3cosx \\ & \rm \: tan3x = \dfrac{3tanx - {tan}^{3} x}{1 - {3tan}^{2}x} \end{array}}} \\ \\[/tex]