Question

sec theta - tan theta = a+1/a-1 ,then cos theta=?​

Answer 1

Step-by-step explanation:

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Answer 2

hey mate,...

here's your answer,...

$[tex] \sec( \alpha ) - \tan( \alpha ) = \frac{a + 1}{a - 1} \\ \frac{1}{ \cos( \alpha ) } - \frac{ \sin( \alpha ) }{ \cos( \alpha ) } = \frac{ {(a + 1)}^{2} }{ {a}^{2} - 1 } \\ = \frac{1 - \sin( \alpha ) }{ \cos( \alpha ) } = \frac{( {1 + a)}^{2} }{ {a}^{2} - 1 } \\ = \frac{1 - \sin( \alpha ) \times 1 + \sin( \alpha ) }{ \cos( \alpha ) \times (1 + \sin( \alpha ) ) } = \frac{ {(1 + a)}^{2} }{ {a}^{2} - 1 } \\ = \frac{ {1 - \sin( \alpha ) }^{2} }{ \cos( \alpha ) \times (1 + \sin( \alpha ) } = \frac{ {(1 + a)}^{2} }{ {a}^{2} - 1 } \\ = \frac{ { \cos( \alpha ) }^{2} }{ \cos( \alpha ) \times (1 + \sin( \alpha ) ) } = \frac{ \cos( \alpha ) }{1 + \sin( \alpha ) } = \frac{ {(1 + a)}^{2} }{ {a}^{2} - 1 } \\ = \frac{ \cos( \alpha ) }{1} = \frac{ {(1 + a)}^{2}(1 + \sin( \alpha ) }{ {a}^{2} - 1 } </p><p></p><p></p><p>$

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