Question

Prove LHS and RHS in 1-cos Ø/sin Ø = sin Ø/1+cos Ø​

Answer 1

Answer:

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.