Question

prove that
sin A÷
sec A + tan A-1+
cos A÷
cosecA+cot A-1=1​

Answer 1

Answer:

sin(a)/{sec(a) + tan(a) - 1} = {sin(a)*cos(a)}/{1 + sin(a) - cos(a)} [Multiplying by cos(a)]

cos(a)/{csc(a) + cot(a) - 1} = {cos(a)*sin(a)}/{1 + cos(a) - sin(a)} [Multiplying by sin(a)]

left side = {sin(a)*cos(a)}/{1 + sin(a) - cos(a)} + {cos(a)*sin(a)}/{1 + cos(a) - sin(a)}

= {cos(a)*sin(a)}[1/{1 + sin(a) - cos(a)} + 1/{1 + cos(a) - sin(a)}]

= {cos(a)*sin(a)}*[{1 + cos(a) - sin(a) + 1 + sin(a) - cos(a)}/{1 + sin(a) - cos(a)*{1 + cos(a) - sin(a)}]

= {cos(a)*sin(a)}*[2/{1 - (sin a - cos a)²} = {cos(a)*sin(a)}*[2/2*sin(a)*cos(a)} = 2sin(a)cos(a)/2sin(a)cos(a) = 1

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Answer 2

i) sin(a)/{sec(a) + tan(a) - 1} = {sin(a)*cos(a)}/{1 + sin(a) - cos(a)} [Multiplying by cos(a)]

ii) cos(a)/{csc(a) + cot(a) - 1} = {cos(a)*sin(a)}/{1 + cos(a) - sin(a)} [Multiplying by sin(a)]

iii) Thus left side = {sin(a)*cos(a)}/{1 + sin(a) - cos(a)} + {cos(a)*sin(a)}/{1 + cos(a) - sin(a)}

= {cos(a)*sin(a)}[1/{1 + sin(a) - cos(a)} + 1/{1 + cos(a) - sin(a)}]

= {cos(a)*sin(a)}*[{1 + cos(a) - sin(a) + 1 + sin(a) - cos(a)}/{1 + sin(a) - cos(a)*{1 + cos(a) - sin(a)}]

= {cos(a)*sin(a)}*[2/{1 - (sin a - cos a)²} = {cos(a)*sin(a)}*[2/2*sin(a)*cos(a)} = 2sin(a)cos(a)/2sin(a)cos(a) = 1

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