Question

Prove that ( sinØ - cos Ø +1 )/(sinØ + cos Ø -1 ) = 1/(SecØ - tan Ø)
Using identity sec² Ø = 1 + tan² Ø

Answer 1

Q. Prove that (sin Ø - cos Ø + 1) / (sin Ø + cos Ø - 1) = 1/(sec Ø - tan Ø) using the identity sec² Ø = 1 + tan² Ø

Solution :

We apply the identity involving sec Ø and tan Ø, let us first convert the LHS in terms of sec Ø and tan Ø by dividing numerator and denominator by cos Ø.

LHS = ( sinØ- cosØ + 1 ) / ( sin Ø + cos Ø - 1 )

= ( tanØ - 1 + secØ ) / ( tanØ + 1 - secØ )

= ( tanØ + secØ ) - 1 / ( tanØ - secØ ) + 1

= { (tanØ + secØ) - 1 }(tanØ - secØ) / (tanØ - secØ) + 1 }(tanØ - secØ)

= [ ( tan²Ø - sec²Ø) - ( tanØ - secØ ) ] / [ { tanØ - secØ + 1 } ( tanØ - secØ ) ]

= ( - 1 - tanØ + secØ ) / ( tanØ - secØ + 1 )(tanØ - secØ)

= -1 / (tanØ - secØ)

= 1/(secØ - tanØ)

Which is the RHS of the identity, we are required to prove.

Answer 2

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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