Answers

2014-09-18T19:28:58+05:30
L am gonna prove these two equations
(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
 
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