Math, asked by aditya99975, 10 months ago

(sin theta - cos theta + 1 ) / ( sin theta + cos theta - 1) = 1 / sec theta - tan theta ​

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Answered by sanjaverma4243
7

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Answered by Anonymous
230

\red\bigstarQUESTION:-

Prove that

 \ \bigstar \sf  \dfrac{sin \theta - cos \theta + 1}{sin \theta + cos  \theta + 1}  =  \dfrac{1}{ \sec \theta  - tan \theta}

\blue\bigstarIDENTITY USED:-

 \boxed{  \pink{\star \:  \sf{ sec}^{2} \theta - {tan}^{2} \theta = 1  }}

\green\bigstarSOLUTION:-

 \sf LHS=  \orange{\dfrac{sin \theta-cos \theta+1}{sin \theta+cos \theta-1} }

 \sf =  \dfrac{ \dfrac{sin \theta}{cos \theta}  - 1 +  \dfrac{1}{cos \theta} }{ \dfrac{sin \theta}{cos \theta} + 1 -  \dfrac{1}{cos \theta}  }  \bigg[on\:Dividing \:  num. \: and \: denom.by \: cos \theta \bigg]

 \sf =  \dfrac{(tan \theta - 1 + sec \theta) }{(tan \theta + 1 -  \sec \theta)}

 \sf =  \dfrac{(sec \theta + tan \theta - 1)}{(tan \theta - sec \theta + 1)}

 =   \sf\dfrac{(sec \theta + tan \theta) - ({sec}^{2} \theta  - {tan}^{2}\theta)}{(tan \theta - sec \theta + 1)}\bigg[\to 1= {sec}^{2}  \theta- {tan}^{2} \theta \bigg]

 =   \sf\dfrac{(sec \theta + tan \theta) [1- ({sec} \theta  - {tan}\theta)]}{(tan \theta - sec \theta + 1)}

 =   \sf\dfrac{(sec \theta + tan \theta) (tan \theta - sec \theta + 1)}{(tan \theta - sec \theta + 1)}

 \sf \red{ = (sec \theta +tan \theta)}

 \sf RHS=  \blue{\dfrac{1}{(sec \theta-tan \theta)} }

 \sf = \dfrac{1}{(sec \theta-tan \theta)} \times  \dfrac{(sec \theta + tan \theta)}{(sec \theta + tan \theta)}

 \sf =  \dfrac{(sec \theta + tan \theta)}{({sec}^{2} -{ tan}^{2} ) }

 \sf \pink {= (sec \theta + tan \theta)}

 \bigstar\sf\orange{LHS}= \blue{RHS}

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