Question

solve this question please​

Answer 1

Step-by-step explanation:

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

Answer 2

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