CBSE BOARD XII, asked by hardesuwa6064, 15 days ago

Solve the equations.tan-1(1-x1+x)=12tan-1x,(x>0)

Answers

Answered by mathdude500
9

Appropriate Question :- Solve for x :-

\sf \:  {tan}^{ - 1}\bigg(\dfrac{1 - x}{1 + x} \bigg)   = \dfrac{1}{2} {tan}^{ - 1}x, \:  \:  \: x  >  0 \\  \\  \\

\large\underline{\sf{Solution-}}

Given inverse trigonometric function is

\sf \:  {tan}^{ - 1}\bigg(\dfrac{1 - x}{1 + x} \bigg)   = \dfrac{1}{2} {tan}^{ - 1}x\\  \\  \\

can be rewritten as

\sf \:  {tan}^{ - 1}\bigg(\dfrac{1 - x}{1 + x \times 1} \bigg)   = \dfrac{1}{2} {tan}^{ - 1}x\\  \\  \\

We know,

\sf \:  \boxed{ \sf{ \:{tan}^{ - 1}\bigg(\dfrac{x - y}{1 + x \times y} \bigg)   =  {tan}^{ - 1}x + {tan}^{ - 1}y \: }} \: \\  \\

So, using this result, we get

\sf \: {tan}^{ - 1}1 - {tan}^{ - 1}x =  \frac{1}{2}{tan}^{ - 1}x \\  \\

\sf \: {tan}^{ - 1}\bigg(tan\dfrac{\pi}{4} \bigg)  =  {tan}^{ - 1}x  +   \frac{1}{2}{tan}^{ - 1}x \\  \\

\sf \: \dfrac{\pi}{4}   =   \frac{3}{2}{tan}^{ - 1}x \\  \\

\sf \: \dfrac{\pi}{6}   =   {tan}^{ - 1}x \\  \\

\sf \: x = tan\dfrac{\pi}{6} \\  \\

\sf \:  \implies \: x =  \frac{1}{ \sqrt{3} }  \\  \\  \\

\rule{190pt}{2pt}

Additional information

\begin{gathered}\boxed{\begin{array}{c|c} \bf Function & \bf Range \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf y =  {sin}^{ - 1}(sinx) & \sf  x \:  \: if -\dfrac{\pi  }{2} \leqslant x \leqslant \dfrac{\pi  }{2}\\ \\ \sf y =  {cos}^{ - 1}(cosx) & \sf x \:  \: if \: 0 \leqslant y \leqslant \pi \\ \\ \sf y =  {tan}^{ - 1}(tanx) & \sf x \:  \: if \:  - \dfrac{\pi  }{2} < x < \dfrac{\pi  }{2}\\ \\ \sf y =  {cosec}^{ - 1}(cosecx) & \sf x \:  \: if \: x \:  \in \: \bigg[ - \dfrac{\pi}{2}, \: \dfrac{\pi  }{2}\bigg] -  \{0 \}\\ \\ \sf y =  {sec}^{ - 1}(secx) & \sf x \:  \: if \: x \:  \in \: [0, \: \pi] \:   -  \: \bigg\{\dfrac{\pi  }{2}\bigg\}\\ \\ \sf y =  {cot}^{ - 1}(cotx) & \sf x \:  \: if \:  \:  \in \: \bigg( -  \dfrac{\pi  }{2} , \dfrac{\pi  }{2}\bigg) -  \{0 \} \end{array}} \\ \end{gathered} \\

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