Question

please solve this question​

Answer 1

Answer:

(1 + sin∅)/cos∅

Step-by-step explanation:

(tan∅ + sec∅ - 1)/(tan∅ - sec∅ + 1)

=> [(tan∅ + sec∅ - (sec²∅ - tan²∅)]/(tan∅ - sec∅ + 1)

=> [(tan∅ + sec∅ - (sec∅ + tan∅)(sec∅ - tan∅)]/(tan∅ - sec∅ + 1)

=> [tan∅ + sec∅(1 - sec∅ + tan∅)]/(tan∅ - sec∅ + 1)

=> tan∅ + sec∅

=> (sin∅/cos∅) + (1/cos∅)

=> (sin∅ + 1)/cos∅

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Answer 2

Question :-

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

Solution :-

$\sf \dfrac{tan \theta + sec \theta - 1}{tan \theta - sec \theta + 1} = \dfrac{1 + sin \theta}{cos \theta}$

Consider LHS

$\sf \dfrac{tan \theta + sec \theta - 1}{tan \theta - sec \theta + 1}$

Substituting 1 = sec² θ - tan² θ in the numerator

$\sf = \dfrac{tan \theta + sec \theta - (sec^{2} \theta - tan^{2} \theta) }{tan \theta - sec \theta + 1}$

Expanding (sec² θ - tan² θ)

$\sf = \dfrac{tan \theta + sec \theta - (sec \theta + tan\theta)(sec \theta - tan \theta) }{tan \theta - sec \theta + 1}$

[ Because, x² - y² = (x + y)(x - y) ]

$\sf = \dfrac{sec \theta + tan \theta - (sec \theta + tan\theta)(sec \theta - tan \theta) }{tan \theta - sec \theta + 1}$

Taking (sec θ + tan θ) common in numerator

$\sf = \dfrac{sec \theta + tan \theta \{1 -(sec \theta - tan \theta) \} }{tan \theta - sec \theta + 1}$

$\sf = \dfrac{sec \theta + tan \theta(1 -sec \theta + tan \theta) }{tan \theta - sec \theta + 1}$

Rearranging the terms in denominator

$\sf = \dfrac{sec \theta + tan \theta( \cancel{1 -sec \theta + tan \theta}) }{ \cancel{1 - sec \theta + tan \theta} }$

$\sf = sec \theta + tan \theta$

$\sf= \dfrac{1}{cos \theta} + \dfrac{sin \theta}{cos \theta}$

[ Because, sec θ = 1/cos θ and tan θ = sin θ/cos θ ]

$\sf= \dfrac{1 + sin \theta}{cos \theta}$

= RHS

∴ LHS = RHS

Hence proved.