Question

if cosec theta+cot theta=p show that sec theta=p^2+1/p^2 -1​

Answer 1

HIIII......{∆}

THERE IS YOUR ANSWER.....{∆}

$question.... \\ \csc( \alpha ) + \cot( \alpha ) = p \\ \\ than \: prove \: that..... \\ \sec( \alpha ) = \frac{ {p}^{2} + 1 }{ {p}^{2} - 1 } \\ solution..... \\ \frac{1}{ \sin( \alpha ) } + \cot( \alpha ) = p \\ \\ 1 + \sin( \alpha ) . \cot( \alpha ) = p \sin( \alpha ) \\ \\ 1 + \cos( \alpha ) = p \sin( \alpha ) \\ \\ {(1 + \cos( \alpha )) }^{2} = {p}^{2} { \sin( \alpha ) }^{2} \\ \\ 1 + { \cos( \alpha ) }^{2} + 2 \cos( \alpha ) = {p}^{2} { \sin( \alpha ) }^{2} \\ \\ 1 + (1 - { \sin( \alpha ) }^{2} ) + 2 \cos( \alpha ) = {p}^{2} { \sin( \alpha ) }^{2} \\ \\ 2(1 + \cos( \alpha ) ) = { \sin( \alpha ) }^{2} (1 + {p}^{2} ) \\ \\ 2(1 + \cos( \alpha )) = 1 - { \cos( \alpha ) }^{2} (1 + {p}^{2} ) \\ \\ \frac{2}{1 + {p}^{2} } = \frac{1 - { \cos( \alpha ) }^{2} }{1 + \cos( \alpha ) } \\ \\ \frac{2}{1 + {p}^{2} } = 1 - \cos( \alpha ) \\ \\ 2 = (1 + {p}^{2} ) - (1 + {p}^{2} ) \cos( \alpha ) \\ \\ 1 - {p}^{2} = - (1 + {p}^{2} ) \cos( \alpha ) \\ \\ \frac{ {p}^{2} - 1 }{ {p}^{2} + 1 } = \cos( \alpha ) \\ \\ \sec( \alpha) = \frac{ {p}^{2} + 1 }{ {p}^{2} - 1 } \\ |bye|$

HOPE IT'S HELP YOU.....{∆}

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