Question

sec(x+a) cosec(x+b)

Answer 1

$\frac{\tan(x+a) + \cot(x+b)}{\cos(b-a)}$

Answer 2

The given expression is $\sec(x+a) \operatorname{cosec}(x+b)$. We can rewrite this in terms of sine and cosine: $E = \frac{1}{\cos(x+a)} \cdot \frac{1}{\sin(x+b)} = \frac{1}{\cos(x+a)\sin(x+b)}$.

To simplify or rewrite this expression, we can use trigonometric identities. Consider the difference of the arguments in the denominator: $(x+b) - (x+a) = b-a$. We can write the numerator '1' using the identity for $\cos(A-B)$. $\cos(b-a) = \cos((x+b) - (x+a)) = \cos(x+b)\cos(x+a) + \sin(x+b)\sin(x+a)$.

Now, let's divide $\cos(b-a)$ by the denominator of the original expression, $\cos(x+a)\sin(x+b)$: $\frac{\cos(b-a)}{\cos(x+a)\sin(x+b)} = \frac{\cos(x+b)\cos(x+a) + \sin(x+b)\sin(x+a)}{\cos(x+a)\sin(x+b)}$ Split the fraction into two terms: $= \frac{\cos(x+b)\cos(x+a)}{\cos(x+a)\sin(x+b)} + \frac{\sin(x+b)\sin(x+a)}{\cos(x+a)\sin(x+b)}$ Cancel the common terms in each fraction: $= \frac{\cos(x+b)}{\sin(x+b)} + \frac{\sin(x+a)}{\cos(x+a)}$ Using the definitions of cotangent and tangent: $= \cot(x+b) + \tan(x+a)$.

So, we have $\frac{\cos(b-a)}{\cos(x+a)\sin(x+b)} = \cot(x+b) + \tan(x+a)$. We want to find the expression for $\frac{1}{\cos(x+a)\sin(x+b)}$. Assuming $\cos(b-a) \neq 0$, we can divide both sides by $\cos(b-a)$: $\frac{1}{\cos(x+a)\sin(x+b)} = \frac{1}{\cos(b-a)} [\cot(x+b) + \tan(x+a)]$.

Thus, $\sec(x+a) \operatorname{cosec}(x+b) = \frac{1}{\cos(b-a)} [\tan(x+a) + \cot(x+b)]$.

This rewrite is valid provided $\cos(b-a) \neq 0$, i.e., $b-a \neq (n + \frac{1}{2})\pi$ for any integer $n$. If $\cos(b-a) = 0$, then $b-a = (n + \frac{1}{2})\pi$. In this case, $\sin(b-a) = \sin((n + \frac{1}{2})\pi) = (-1)^n \neq 0$. In this specific case, $b = a + (n + \frac{1}{2})\pi$. $\sin(x+b) = \sin(x+a + (n + \frac{1}{2})\pi) = \sin((x+a) + \frac{\pi}{2} + n\pi) = \cos(x+a + n\pi) = \cos(x+a)\cos(n\pi) - \sin(x+a)\sin(n\pi) = \cos(x+a)(-1)^n$. The expression becomes $\sec(x+a) \operatorname{cosec}(x+b) = \frac{1}{\cos(x+a)} \frac{1}{(-1)^n \cos(x+a)} = \frac{(-1)^n}{\cos^2(x+a)} = (-1)^n \sec^2(x+a)$.

However, the form involving $\tan$ and $\cot$ is the standard way to rewrite this expression in the general case.