Math, asked by subhashattri07, 5 months ago

solve this question please​

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Answered by Anonymous
1

Step-by-step explanation:

here is your answer of your question hope it will help !

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Answered by sandy1816
4

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using \:  \: tan3 \alpha  =  \frac{3tan \alpha  -  {tan}^{3}  \alpha }{1 - 3 {tan}^{2}  \alpha }  \\   \:  \: 2 {tan}^{ - 1}  \alpha  =tan ^{ - 1}   \frac{ {2 \alpha } }{1 -  { \alpha }^{2} }  \\  \\ \\ let \:  \: x = tan \theta \:  \: then \:   \: \theta =  {tan}^{ - 1}   x \\ rhs \:  \:  \:  {tan}^{ - 1} ( \frac{3x -  {x}^{3} }{1 - 3 {x}^{2} } ) \\  =  {tan}^{ - 1} ( \frac{3tan \theta -  {tan}^{3}  \theta  }{1 -  3{tan}^{2} \theta  }  \: ) \\  =  {tan}^{ - 1} (tan3 \theta ) \\  = 3 \theta \\  = 3 {tan}^{ - 1}  x \\  =  {tan}^{ - 1} x + 2 {tan}^{ - 1} x \\  =  {tan}^{ - 1}  +  {tan}^{ - 1}  \frac{2x}{1 -  {x}^{2} }  \\  =l hs

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