Question

i ) $\frac{tan³∅ - 1}{tan∅ - 1 } = sec²∅ + tan∅$
ii ) $\frac{sin∅-cos∅+1}{sin∅-cos∅ - 1} = \frac{1}{sec∅ \: - tan \: ∅}$
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Answer 1

Step-by-step explanation:

(ii)L.H.S

=

sinθ+cosθ−1

sinθ−cosθ+1

=

tanθ−secθ+1

tanθ+secθ−1

=

tanθ−secθ+1

(tanθ+secθ)−(sec

2

θ−tan

2

θ)

=

tanθ−secθ+1

(tanθ+secθ)−(secθ−tanθ)(secθ+tanθ)

=

(1−secθ+tanθ)

(tanθ+secθ)(1−secθ+tanθ)

=secθ+tanθ

Multiplying and Dividing by secθ−tanθ we get,

=

secθ−tanθ

sec

2

θ−tan

2

θ

=

secθ−tanθ

1