Proving \frac{1+\csc^2 A \tan^2 C}{1+\csc^2 B \tan^2 C}=\frac{1+\cot^2 A \sin^2

Seamus Kent

Seamus Kent

Answered question

2022-01-24

Proving 1+csc2Atan2C1+csc2Btan2C=1+cot2Asin2C1+cot2Bsin2C

Answer & Explanation

Karly Logan

Karly Logan

Beginner2022-01-25Added 11 answers

hose to manipulate the left hand side of the equation, by firstly replacing cot2A with csc2A1 according to the identities. After doing the same with the denominator,
RHS=1+(csc2A1)sin2C1+(csc2B1)sin2C=1+csc2Asin2Csin2C1+csc2Bsin2Csin2C
1+csc2Asin2Csin2C1+csc2Bsin2Csin2C=(cos2C+csc2Asin2C)sec2C(cos2C+csc2Bsin2C)sec2C=1+csc2Atan2C1+csc2Btan2C

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