I need help proving this identity
cscx + secx = (cosx + sinx)(secx)(cscx)
LS = 1/sinx + 1/cosx
= (cosx + sinx)/(sinxcosx)
= (cosx + sinx)(1/sinx)(1/cosx)
= (cosx+ sinx)(cscx)(sec)
= RS
To prove the given identity: cscx + secx = (cosx + sinx)(secx)(cscx), we'll simplify the left-hand side and the right-hand side separately, and then check if they are equal.
Let's start by simplifying the left-hand side of the equation:
1. We know that cscx is the reciprocal of sinx: cscx = 1/sinx.
2. Similarly, secx is the reciprocal of cosx: secx = 1/cosx.
Substituting these values, the left-hand side becomes:
(1/sinx) + (1/cosx)
To add these fractions, we need a common denominator. The least common denominator is sinx * cosx. So, multiply the first term by cosx/cosx and the second term by sinx/sinx:
[(1 * cosx) / (sinx * cosx)] + [(1 * sinx) / (sinx * cosx)]
(cosx + sinx) / (sinx * cosx)
Now, let's simplify the right-hand side of the equation:
(cosx + sinx)(secx)(cscx)
Since secx is equal to 1/cosx, and cscx is equal to 1/sinx, we can substitute these values into the equation:
(cosx + sinx)(1/cosx)(1/sinx)
Multiplying the numerators together and multiplying the denominators together, we get:
[(cosx + sinx) / (sinx * cosx)]
Comparing the simplified left-hand side and right-hand side, we can see that they are equal:
(cosx + sinx) / (sinx * cosx) = (cosx + sinx) / (sinx * cosx)
Hence, the identity cscx + secx = (cosx + sinx)(secx)(cscx) is proven.
Keep in mind that when proving trigonometric identities, it's important to simplify both sides of the equation using known trigonometric ratios and algebraic manipulations.