Question

Show that sin?-cos?+1/sin?+cos?-1= 1/sec?-tan?.

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Answer 1

GIVEN :

(sin θ – cos θ + 1)/(sin θ + cos θ – 1)

SOLUTION :

LHS: (sin θ – cos θ + 1)/(sin θ + cos θ – 1)

•Divide by cos θ in numerator and denominator,

={(sin θ – cos θ + 1)/cos θ}/{(sin θ + cos θ – 1)/cos θ}

=(sin θ /cos θ – cos θ /cos θ + 1/cos θ)/(sin θ / cos θ + cos θ /cos θ – 1/cos θ)

=(tan θ – 1 + sec θ)/(tan θ + 1 – sec θ)

= (sec θ + tan θ – 1)/(tan θ + 1 – sec θ)

={sec θ + tan θ – (sec² θ – tan² θ)}/(tan θ + 1 – sec θ) [Since sec² θ – tan²θ = 1]

={(sec θ + tan θ) – (sec θ – tan θ) * (sec θ + tan θ)}/(tan θ + 1 – sec θ)

=[(sec θ + tan θ) * {1 – (sec θ – tan θ)}]/(tan θ + 1 – sec θ)

=[(sec θ + tan θ) * (1 – sec θ + tan θ)}]/(tan θ + 1 – sec θ)

= sec θ + tan θ

Now, rationalize it.

=(sec θ + tan θ)*(sec θ – tan θ)/(sec θ – tan θ)

= (sec² θ – tan² θ)/(sec θ – tan θ)

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

RHS .

Hence, (sin θ – cos θ + 1)/(sin θ + cos θ – 1) = 1/(sec θ – tan θ) .