(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
SOLUTION
TO PROVE
PROOF
LHS
Dividing numerator and denominator both by cos θ we get
= RHS
Hence proved
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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