Question

Prove that : Sin θ - cos θ + 1/ sin θ + cos θ - 1 = 1/sec θ - Tan θ​

Answer 1

Answer:

Consider the L.H.S

sinθ+cosθ−1

sinθ−cosθ+1

=(

sinθ+cosθ−1

sinθ−cosθ+1

)×(

sinθ+cosθ+1

sinθ+cosθ+1

)

=(

sinθ+cosθ−1

sinθ+1−cosθ

)×(

sinθ+cosθ+1

sinθ+1+cosθ

)

=

(sinθ+cosθ)

2

−1

2

(sinθ+1)

2

−cos

2

θ

=

sin

2

θ+cos

2

θ+2sinθcosθ−1

sin

2

θ+1+2sinθ−cos

2

θ

Since, sin

2

θ+cos

2

θ=1

Therefore,

=

1+2sinθcosθ−1

1−cos

2

θ+1+2sinθ−cos

2

θ

=

2sinθcosθ

2−2cos

2

θ+2sinθ

=

sinθcosθ

1−cos

2

θ+sinθ

=

sinθcosθ

sin

2

θ+sinθ

=

cosθ

sinθ+1

=

cosθ

1

+

cosθ

sinθ

=secθ+tanθ

=(secθ+tanθ)×(

secθ−tanθ

secθ−tanθ

)

=

secθ−tanθ

sec

2

θ−tan

2

θ

We know that

sec

2

θ−tan

2

θ=1

Therefore,

=

secθ−tanθ

1

Hence, proved

Answer 2

Answer:

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