MHB Prove identity (sinx+cosx)/(secx+cscx)= sinxcosx

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To prove the identity (sinx + cosx)/(secx + cscx) = sinx cosx, start by rewriting the left-hand side in terms of sine and cosine functions. The expression becomes (sinx + cosx)/(1/cosx + 1/sinx). Next, combine the terms in the denominator to simplify the expression further. This approach leads to a clearer path for demonstrating the equality with the right-hand side. The proof ultimately hinges on manipulating the trigonometric identities effectively.
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(sinx+cosx)/(secx+cscx)= sinxcosx if you could list out the steps it would be appreciated
 
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Hello, and welcome to MHB! (Wave)

Since the RHS is in terms of the sine and cosine functions, the first thing I would do is write the LHS in terms of these functions only:

$$\frac{\sin(x)+\cos(x)}{\dfrac{1}{\cos(x)}+\dfrac{1}{\sin(x)}}$$

Now, combine terms in the denominator...what do you get?
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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