Trigonometric Equations: Prove these identities

1) (cosx/1 - tanx) + (sinx/1 - cotx) = sinx + cosx

2) (tanx + cotx - 1)(sinx + cosx) = (secx/cosec^2x) + (cosecx/sec^2x)


Please show working when proving. (Recommended to use the Left hand side/Right hand side proof method)

Thanks in advance.

"/"=Divide (vinculum)

2)

... LHS

= ( tan x + cot x - 1 )( sin x + cos x )

= [ ( sin x / cos x ) + ( cos x / sin x ) - 1 ] ( sin x + cos x )

= { [ ( sin² x + cos² x ) - sin x cos x ] / ( sin x cos x ) } ( sin x + cos x )

= ( sin x + cos x )( sin² x - sin x cos x + cos² x ) / ( sin x cos x )

= ( sin³ x + cos³ x ) / ( sin x cos x ) ...... Now Split

= ( sin² x / cos x ) + ( cos² x / sin x )

= [ ( 1 / cos x ) / ( 1 / sin² x ) ] + [ ( 1 / sin x ) / ( 1 / cos² x ) ]

= ( sec x / csc² x ) + ( csc x / sec² x )

= RHS

Hence, the result.
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Iceman got there first.
So the BA belongs to him!

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1)
LHS = cosx/(1 - sinx/cosx) + sinx/(1 - cosx/sinx)
= cosx/(cosx - sinx)/cosx + sinx/(sinx - cosx)/sinx
= cos^2x/(cosx - sinx) - sin^2x/(cosx - sinx)
= [(cosx - sinx)(cosx + sinx)]/(cosx - sinx)
= sinx + cosx = RHS