Question

4x secA = 1+ 4x² হয তবে পমান করো secA + tanA = 2x ​

Answer 1

Answer:

$\sec( \alpha ) + \tan( \alpha ) = 2x$

Step-by-step explanation:

$\sec( \alpha ) = \frac{1 + 4 {x}^{2} }{4x} \\ \\ \tan( \alpha ) = \sqrt{ { \sec( \alpha ) }^{2} - 1} \\ \\ \tan( \alpha ) = \sqrt{ {( \frac{1 + 4 {x}^{2} }{4x}) }^{2} - 1 } \\ \\ = \frac{ \sqrt{ {(1 + 4 {x}^{2} )}^{2} - {(4x)}^{2} } }{4x} \\ \\ = \frac{ \sqrt{ {(4 {x}^{2} + 1 - 4x) }^{}(4 {x}^{2} + 1 + 4x) } }{4x} \\ \\ = \frac{(2 {x}^{} - 1) (2x + 1)}{4x} \\ \\ therefore \: \: \: \\ \\ \sec( \alpha ) + \tan( \alpha ) = \frac{1 + 4 {x}^{2} }{4x} + \frac{(2 {x}^{} - 1)(2x + 1) }{4x} \\ \\ = \frac{8 {x}^{2} }{4x} \\ \\ = 2x \\ \\ = rhs$