Question

TanA= 2x(x+1)/2x+1 find sinA and cosA

Answer 1

$\huge\color{Black}\boxed{\colorbox{yellow}{❀Answer❀}}$

$tanA= \frac{2x(x + 1)}{2x + 1} \\ secA = \sqrt{1 + {tan}^{2}A } \\ = \sqrt{1 + \frac{4 {x}^{2} ({x + 1})^{2} }{( {2x + 1})^{2} } } \\ = \sqrt{ \frac{ ({2x + 1})^{2} + 4 {x}^{2}( {x + 1})^{2} }{ ({2x + 1})^{2} } } \\ = \sqrt{ \frac{4 {x}^{2} + 4x + 1 + 4 {x}^{2} ( { {x}^{2} + 2x + 1}) }{( {2x + 1})^{2} } } \\ = \sqrt{ \frac{4 {x}^{2} + 4x + 1 + 4 {x}^{4} + 8 {x}^{3} + 4 {x}^{2} }{( {2x + 1})^{2} } } \\ = \sqrt{ \frac{4 {x}^{4} + 4 {x}^{2} + 1 + 2(4 {x}^{3} + 2x + 2 {x}^{2}) }{ ({2x + 1})^{2} } } \\ = \sqrt{ \frac{ ({2 {x}^{2} + 2x + 1 })^{2} }{ ({2x + 1})^{2} } } \\ = \frac{2 {x}^{2} + 2x + 1}{ 2x + 1 } \\ cosA = \frac{2x + 1}{2 {x}^{2} + 2x + 1} \\ sinA = \sqrt{1 - {cos}^{2} A} \\ = \sqrt{1 - ( { \frac{2x + 1}{2 {x}^{2} + 2x + 1} })^{2} } \\ = \sqrt{ \frac{ ({2 {x}^{2} + 2x + 1})^{2} - ( {2x + 1})^{2} }{( {2 {x}^{2} + 2x + 1 })^{2} } } \\ = \sqrt{ \frac{(2 {x}^{2} + 2x + 1 + 2x + 1)(2 {x}^{2} + 2x + 1 - 2x - 1 )}{ ({2 {x}^{2} + 2x + 1})^{2} } } \\ = \sqrt{ \frac{(2 {x}^{2} + 4x + 1)(2 {x}^{2}) }{( {2 {x}^{2} + 2x + 1 })^{2} } } \\ = \sqrt{ \frac{( {x}^{2} + 2x + 1)(4 {x}^{2}) }{( { 2 {x}^{2} + 2x + 1})^{2} } } \\ = \frac{(x + 1)2x}{2 {x}^{2} + 2x + 1 }$