Question

Prove that :-
(tan θ + sec θ - 1)/(tan θ - sec θ + 1) = (1 + sin θ)/cos θ

Answer 1

L.H.S = (tan θ + sec θ - 1)/(tan θ - sec θ + 1)

= [(tan θ + sec θ) - (sec²θ - tan²θ)]/(tan θ - sec θ + 1)

• [» sec²θ - tan²θ = 1]

= {(tanθ+secθ) - (secθ+tanθ) (secθ-tanθ)}/(tanθ-sec θ + 1)

= {(tan θ + sec θ) (1 - sec θ + tan θ)}/(tan θ - sec θ + 1)

= {(tan θ + sec θ) (tan θ - sec θ + 1)}/(tan θ - sec θ + 1)

= tan θ + sec θ

= (sin θ/cos θ) + (1/cos θ)

= (sin θ + 1)/cos θ

= (1 + sin θ)/cos θ = R.H.S Proved.

Answer 2

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${\huge{\bold{\underline{Answer:-}}}}$

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Here, i am holding the "THETAS" as "A".

So, we can proceed our solution as under

= $\huge\ { \fbox{ \tt{ L.H.S }}}$

(tan A + sec A - 1)/(tan A - sec A + 1)

= (tan A + sec A - sec²A + tan²A)/(tan A - sec A + 1)

= [tan A + sec A - {(sec A+tan A) (sec A - tan A)}]/[tan A - sec A + 1]

= [tan A + sec A (1 - sec A + tan A)]/(tan A - sec A + 1)

= tan A + sec A

= sin A/cos A + 1/cos A

= ( 1 + sin A ) / cos A

= $\huge\ { \fbox{ \tt{ R.H.S }}}$

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ULTIMATELY,

L.H.S. = R.H.S. (PROVED)

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