Math, asked by suramanksc2017, 11 months ago

(sin 0 - cos 0 +1)/(sin 0 + cos 0 -1) = 1/(sec 0 - tan 0)

Here 0 is theta. Don't assume it as zero.​

Answers

Answered by pulakmath007
17

SOLUTION

TO PROVE

 \displaystyle \sf{ \frac{ \sin \theta -  \cos \theta   + 1}{\sin \theta  +   \cos \theta   -  1} =  \frac{1}{ \sec \theta \:  -  \tan \theta}  }

PROOF

LHS

 \displaystyle \sf{ =  \frac{ \sin \theta -  \cos \theta   + 1}{\sin \theta  +   \cos \theta   -  1}  }

Dividing numerator and denominator both by cos θ we get

 \displaystyle \sf{ =  \frac{ \tan \theta - 1 +  \sec \theta  }{\tan \theta  +  1 -  \sec \theta   }  }

 \displaystyle \sf{ =  \frac{ \tan \theta - 1 +  \sec \theta  }{\tan \theta   -  \sec \theta + 1   }  }

 \displaystyle \sf{ =  \frac{ \tan \theta - 1 +  \sec \theta  }{(\tan \theta   -  \sec \theta )  - ( { \tan}^{2}  \theta -  { \sec}^{2} \theta )  }  }

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

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

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

 \displaystyle \sf{ =  \frac{1  }{\sec \theta - \tan \theta  }  }

= RHS

Hence proved

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Answered by daksh8785
3

Answer:

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Step-by-step explanation:

TO PROVE

\displaystyle \sf{ \frac{ \sin \theta - \cos \theta + 1}{\sin \theta + \cos \theta - 1} = \frac{1}{ \sec \theta \: - \tan \theta} }

sinθ+cosθ−1

sinθ−cosθ+1

=

secθ−tanθ

1

PROOF

LHS

\displaystyle \sf{ = \frac{ \sin \theta - \cos \theta + 1}{\sin \theta + \cos \theta - 1} }=

sinθ+cosθ−1

sinθ−cosθ+1

Dividing numerator and denominator both by cos θ we get

\displaystyle \sf{ = \frac{ \tan \theta - 1 + \sec \theta }{\tan \theta + 1 - \sec \theta } }=

tanθ+1−secθ

tanθ−1+secθ

\displaystyle \sf{ = \frac{ \tan \theta - 1 + \sec \theta }{\tan \theta - \sec \theta + 1 } }=

tanθ−secθ+1

tanθ−1+secθ

\displaystyle \sf{ = \frac{ \tan \theta - 1 + \sec \theta }{(\tan \theta - \sec \theta ) - ( { \tan}^{2} \theta - { \sec}^{2} \theta ) } }=

(tanθ−secθ)−(tan

2

θ−sec

2

θ)

tanθ−1+secθ

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

(tanθ−secθ)−(tanθ+secθ)(tanθ−secθ)

tanθ−1+secθ

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

(tanθ−secθ)(1−tanθ−secθ)

tanθ−1+secθ

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

(secθ−tanθ)(tanθ+secθ−1)

tanθ+secθ−1

\displaystyle \sf{ = \frac{1 }{\sec \theta - \tan \theta } }=

secθ−tanθ

1

= RHS

Hence proved

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