Question

one upon sec A + tan A minus one upon cos A is equals to one upon Cos A minus one upon sec A - tanA​

Answer 1

Answer:

hey

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

Answer 2

Step-by-step explanation:

your answer is in the linked image

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