Question

1+secØ-tanØ/1+secØ+tanØ=1-sinØ/cosØ

Answer 1

Heya !!

Here's your solution !!!

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LHS = sin∅ - cos∅ +1/ sin∅+cos∅-1

Divided both numerators and denominator by cos∅

LHS = (tan∅ - 1+ sec∅)/(tan∅+1-sec∅)

NOW,

sec²∅ = 1+tan²∅

sec²∅ = tan²∅ = 1

Using above relation at denominator of LHS

LHS = (tan∅ - 1 + sec∅) (tan∅ - sec∅+ sec²∅ - tan²∅)

LHS = ( tan∅ - 1 +sec∅)/

((sec∅ - tan∅)(-1 + sec∅ + tan∅))

LHS = 1/(sec∅ - tan∅)

LHS = RHS

HENCE PROVED

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