(secx - tanx)(cscx+1) =cotx
show me what you did so far, I hope you started on the LS
or at least look at the solution which Reiny already provided for you ...
To prove that (secx - tanx)(cscx+1) = cotx, we need to simplify both sides of the equation and show that they are equal.
Let's start by simplifying the left side of the equation:
(secx - tanx)(cscx+1) = (1/cosx - sinx/cosx)(1/sinx + 1)
= [(1 - sinx)/cosx] * [1/(sinx * cosx)]
= (1 - sinx)/(sinx * cosx * cosx)
= (1 - sinx)/(sinx * cos^2x)
Now, let's simplify the right side of the equation:
cotx = cosx/sinx
To make the two sides equal, we need to simplify (1 - sinx)/(sinx * cos^2x) so that it becomes equal to cosx/sinx.
Let's manipulate the expression (1 - sinx)/(sinx * cos^2x) to match cosx/sinx:
(1 - sinx)/(sinx * cos^2x) = (1 - sinx)/(sinx * cosx * cosx)
Using the trigonometric identity tan^2x + 1 = sec^2x, we can manipulate the expression further:
= (1 - sinx)/(sinx * (1 + tan^2x))
= (1 - sinx)/(sinx + sinx * tan^2x)
= (1 - sinx)/(sinx (1 + tan^2x))
= (1 - sinx)/(sinx * sec^2x)
= (1 - sinx)/(sinx * (1/cos^2x))
= (1 - sinx)/sinx * cos^2x
Now we can see that (1 - sinx)/sinx * cos^2x is equal to cosx/sinx, which matches the right side of the equation.
Therefore, we have shown that (secx - tanx)(cscx+1) = cotx.