prove that 1/secx-tanx -1/cosx =1/cosx-1/secx+tanx
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Answered by
7
Hi ,
LHS = 1/( secx - tanx ) - 1/cos x
= (secx + tanx )/[(secx-tanx)(secx+tanx) -1/cosx
= ( secx + tanx ) / ( sec²x - tan² x ) - secx
[ since sec² x - tan² x = 1 ]
= secx + tanx - secx
= tanx ----( 1 )
RHS = 1/cosx - 1/ ( secx + tanx )
= secx - (secx - tanx ) / [(secx + tanx)(secx-tanx)
= secx - ( secx - tanx ) / ( sec² x - tan² x )
= secx - secx + tanx
= tanx ---( 2 )
from ( 1 ) and ( 2 ) ,
LHS = RHS
I hope this helps you.
: )
LHS = 1/( secx - tanx ) - 1/cos x
= (secx + tanx )/[(secx-tanx)(secx+tanx) -1/cosx
= ( secx + tanx ) / ( sec²x - tan² x ) - secx
[ since sec² x - tan² x = 1 ]
= secx + tanx - secx
= tanx ----( 1 )
RHS = 1/cosx - 1/ ( secx + tanx )
= secx - (secx - tanx ) / [(secx + tanx)(secx-tanx)
= secx - ( secx - tanx ) / ( sec² x - tan² x )
= secx - secx + tanx
= tanx ---( 2 )
from ( 1 ) and ( 2 ) ,
LHS = RHS
I hope this helps you.
: )
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