How do find the value of α in
sec(2α+6°)cos(5α+3°)=1
Use the sum formula for cosine
$\displaystyle \sec{(2a + 6^\circ)}\cos{(5a + 3^\circ)} = 1$
$\displaystyle \frac{\cos{(5a + 3^\circ)}}{\cos{(2a + 6^\circ)}} = 1$
$\displaystyle \frac{\cos{5a}\cos{3^\circ} - \sin{5a}\sin{3^\circ}}{\cos{2a}\cos{6^\circ} - \sin{2a}\sin{6^\circ}} = 1$.
Can you go from here?
Actually I've thought of an easier way...
You have
$\displaystyle \frac{\cos{(5a + 3^\circ)}}{\cos{(2a + 6^\circ)}} = 1$
$\displaystyle \cos{(5a + 3^\circ)} = \cos{(2a + 6^\circ)}$
$\displaystyle 5a + 3^\circ = 2a + 6^\circ$
$\displaystyle 3a = 3^\circ$
$\displaystyle a = 1^\circ$.
Dear somanyquestions,
$\displaystyle sec(2a+6)^{o}cos(5a+3)^{o}=1$
$\displaystyle \frac{\cos{(5a + 3)}^o}{\cos{(2a + 6)^o}} = 1$
$\displaystyle cos(5a+3) = cos(2a+6)$
$\displaystyle \frac{\pi(5a+3)}{180}=2n{\pi}{\pm}\frac{{\pi}(2a+6 )}{180}$ ; $\displaystyle n\in{Z}$
By simplification you would get, $\displaystyle a = \frac{(360n-9)}{7}$ or $\displaystyle a = 120n+1$
Dear Prove It,
Your answer is a particular solution. For example $\displaystyle a = 121^o$ also satisfies the given trignometric equation.
Hope this helps.