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## Answers

(cosecФ - sinФ) (secФ - cosФ)= sinФcosФ and

sinФ cosФ=1/tanФ+cosФ

here goes the proof of first equation

squaring on both sides we get

(cosecФ - sinФ)²(secФ -cosФ)² = sin²Ф cos²Ф

here we are gonna prove LHS(left hand side)=RHS(right hand side)

applying the formula (a-b)² = a² -2ab + b² we get

#(cosec²Ф -2cosecФ sinФ + sin²Ф)(sec²Ф -2 secФ cosФ + cos²Ф)

but cosecФ=1/sinФ and also secФ=1/cosФ applying them to the above equation we get

(cosec²Ф-2 +sin²Ф)(sec²Ф - 2 + cos²Ф)

now if we open the brackets we get

#cosec²Фsec²Ф - 2cosec²Ф + cosec²Фcos²Ф - 2 sec²Ф + 4 - 2 cos²Ф + sin²Фsec²Ф - 2sin²Ф+ sin²Фcos²Ф

#(1/sin²Ф)(1/cos²Ф) - cosec²Ф - cosec²Ф + cos²Ф/sin²Ф - sec²Ф - sec²Ф + 4 - 2 cos²Ф - 2 sin²Ф + sin²Ф/cos²Ф + sin²Фcos²Ф

#1/sin²Фcos²Ф - 1/sin²Ф - cosec²Ф + cot²Ф - 1/cos²Ф - sec²Ф + 4 - 2(cos²Ф+sin²Ф) + tan²Ф + sin²Фcos²Ф

# 1/sin²Фcos²Ф - 1/sin²Ф - 1/cos²Ф - (cosec²Ф - cot²Ф) - (sec²Ф - tan²Ф) + 4 -2(cos²Ф + sin²Ф) + sin²Фcos²Ф

#(1- cos²Ф - sin²Ф/sin²Фcos²Ф) - 1 - 1 + 4 - 2 + sin²Фcos²Ф

#(1 - (cos²Ф + sin²Ф)/sin²Фcos²Ф) + sin²Фcos²Ф

# 1 -1/sin²Фcos²Ф + sin²Фcos²Ф

# sin²Фcos²Ф

LHS =RHS

Hence proved