Question

prove that cos0-si 0+1/ cos0+sin0-1 = 1+cos0/sin0​

Answer 1

Answer:

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Answer 2

Answer:

(cosA-sinA+1)/( cosA+sinA-1) = (1+cosA)/sinA​

Step-by-step explanation:

LHS = (cosA-sinA+1)/(cosA+sinA-1)

Multiplying & dividing by SinA, we get

LHS = SinA (cosA-sinA+1)/SinA (cosA+sinA-1)

(Now, all changes are going to take place in the numerator only.)

={sinAcosA-sin2A+sinA}/sinA (cosA+sinA-1)

={sinAcosA+sinA-sin²A}/sinA (cosA+sinA-1)

={sinAcosA+sinA-(1-cos²A)}/sinA (cosA+sinA-1)

∵sin²A=1-cos²A

=[sinA(cosA+1) - (1-cosA)(1+cosA)]/sinA (cosA+sinA-1)

=(1+cosA)(sinA+cosA-1)/sinA (cosA+sinA-1)

cancelling (cosA+sinA-1) from both numerator  and denominator

=(1+cosA)/sinA

=RHS