prove that 1/ secA+tanA - 1/cosA = 1/cosA - 1/ secA- tanA

LHS:-  [ 1/secA + tanA ] - 1/cosA

=[ 1 (secA-tanA) / secA+tanA (secA-tanA) ]  - secA

= [ secA - tanA / sec2A - tan2A ] -secA

=secA - tanA - secA

= -  tanA

RHS:-  1/cosA - [ 1/secA - tanA ]

= secA - [1 (secA + tanA) / secA - tanA (secA + tanA) ]

= secA - (secA + tanA / sec2A - tan2A ]

= secA - ( secA + tanA )

= secA - secA - tanA

=  - tanA

SO, LHS=RHS. Hence, proved.

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