one upon sec a + tan a minus one upon cos equals to one upon Cos A minus one upon sec a - 10 prove it
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Answer:
LHS=1/sec A+ tan A minus 1/cos A
= 1/ (1/cos A + sinA/cos A) minus 1/cos A
=1/(1+ sinA/cosA) minus 1/cosA
= cosA/1+sinA minus 1/cos A-sinA
= (cos^2A -1-sinA)/cosA (1+sinA)
=(1-sin^2A-1-sinA)/cosA(1+sinA)
=-sinA(sinA+1)/cosA(1+sinA)
=- sinA/cosA
=-tanA
RHS=1/cosA minus 1/(secA-tanA)
=1/cosA minus 1/(1/cosA - sinA/cosA)
=1/cosA minus 1/(1-sinA/cosA)
=1/cosA minus cosA/(1-sinA)
=(1-sinA-cos^2A)/cosA(1-sinA)
=1-sinA-(1-sin^2A)/cosA(1-sinA)
= (sin^2A-sinA)/cosA(1-sinA)
=-sinA(-sinA+1)/cosA(1-sinA)
=-sinA/cosA
=-tanA
=LHS
hence proved
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