Math, asked by shivakrish, 11 months ago

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

Answers

Answered by anu24239
14

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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