If \(\frac{{\sec θ + \tan θ }}{{\sec θ - \tan θ }} = \frac{5}{3}\) , 0° < θ < 90°, then what is the value of (cosec θ + cos θ + cot θ)?
The correct answer is \frac{{16 + 5\sqrt {15} }}{4}.
If you have a proportion \frac{a}{b} = \frac{c}{d}, it implies that \frac{a+b}{b} = \frac{c+d}{d} (Componendo) and \frac{a-b}{b} = \frac{c-d}{d} (Dividendo). Combining these by dividing the Componendo result by the Dividendo result gives \frac{(a+b)/b}{(a-b)/b} = \frac{(c+d)/d}{(c-d)/d}, which simplifies to \frac{a+b}{a-b} = \frac{c+d}{c-d}. This rule simplifies expressions like the one given in the problem statement significantly, often reducing the number of steps needed to find the value of individual trigonometric functions.
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