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prove the following identity showing all steps:
[tex]\frac{cos(x+30)-sin(x+60)}{sin(x)cos(x)} =-sec(x)[/tex]

please help fast


Sagot :

Answer:

See solution below

Step-by-step explanation:

Given the expression

[tex]\frac{cos(x+30)-sin(x+60)}{sin(x)cos(x)} \\[/tex]

Recall that

cos x  = sin(90-x)

cos(x+30 ) = sin (90-(x+30)

= sin(90-x-30)

= sin(60-x)

Substitute

[tex]\frac{sin(60-x)-sin(x+60)}{sin(x)cos(x)} \\= \frac{sin60cosx-cos60sinx)-sinxcos60-cosxsin60)}{sin(x)cos(x)} \\= \frac{-2cos60sinx)}{sin(x)cos(x)} \\= \frac{-2(1/2)sinx)}{sin(x)cos(x)} \\= \frac{-1}{cos(x)}\\= \frac{1}{cos(x)}\\ \\= -sec(x) Proved[/tex]