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honeybear101 · Oct 7, 2010, 07:33 PM

I'm having trouble solving this problem...
sec(x+y)=sec(x)sec(y)/1-tan(x)tan(y)

Unknown008 · Oct 8, 2010, 12:33 AM

What is sec(x+y) ?

Convert into cos:

$sec(x+y) = \frac{1}{cos(x+y)}$

Then, apply the compound angle formula:

$cos(x+y) = cos(x)cos(y) - sin(x)sin(y)$

Replace that in the previous expression:

$\frac{1}{cos(x+y)} = \frac{1}{cos(x)cos(y) - sin(x)sin(y)}$

Now, use something similar to rationalisation, but with a specific function.

$\frac{1}{cos(x)cos(y) - sin(x)sin(y)} = \frac{1}{cos(x)cos(y) - sin(x)sin(y)} \times \frac{\frac{1}{cos(x)cos(y)}}{\frac{1}{cos(x)cos(y )}}$

Let's see where it goes.