Question

(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.​

Answer 1

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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Answer 2

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