Simplifying Trigonometric Expression M for Acute Angle A

The question asks for the value of the expression:

$M = \sqrt{\frac{1+\cos A}{1-\cos A}} - \text{cosec} A - \cot A$

Given that A is an acute angle ($0 < A < 90^\circ$), we need to simplify M.

Simplifying the Square Root Term in M

Let's focus on simplifying the first term, $\sqrt{\frac{1+\cos A}{1-\cos A}}$:

1. To simplify the fraction inside the square root, we can multiply the numerator and denominator by $(1+\cos A)$: $ \sqrt{\frac{1+\cos A}{1-\cos A}} \times \sqrt{\frac{1+\cos A}{1+\cos A}} = \sqrt{\frac{(1+\cos A)^2}{(1-\cos A)(1+\cos A)}} $
2. The denominator $(1-\cos A)(1+\cos A)$ simplifies using the difference of squares formula $(a-b)(a+b) = a^2 - b^2$: $ (1-\cos A)(1+\cos A) = 1^2 - \cos^2 A = 1 - \cos^2 A $
3. Using the fundamental Pythagorean identity $\sin^2 A + \cos^2 A = 1$, we know that $1 - \cos^2 A = \sin^2 A$. Substituting this: $ \sqrt{\frac{(1+\cos A)^2}{\sin^2 A}} $
4. Now, taking the square root of the numerator and the denominator: $ \frac{\sqrt{(1+\cos A)^2}}{\sqrt{\sin^2 A}} = \frac{|1+\cos A|}{|\sin A|} $
5. Since A is an acute angle ($0 < A < 90^\circ$), both $\cos A$ and $\sin A$ are positive.
   • Thus, $1+\cos A$ is positive, so $|1+\cos A| = 1+\cos A$.
   • And $\sin A$ is positive, so $|\sin A| = \sin A$.
   The expression becomes: $ \frac{1+\cos A}{\sin A} $
6. We can split this fraction into two terms: $ \frac{1}{\sin A} + \frac{\cos A}{\sin A} $
7. Recognizing the definitions of cosecant and cotangent ($\text{cosec} A = \frac{1}{\sin A}$ and $\cot A = \frac{\cos A}{\sin A}$), the simplified term is: $ \text{cosec} A + \cot A $

Substituting Back to Find Value of M

Now we substitute the simplified square root term back into the original expression for M:

$M = (\text{cosec} A + \cot A) - \text{cosec} A - \cot A$

By rearranging and combining like terms:

$M = \text{cosec} A - \text{cosec} A + \cot A - \cot A$

$M = 0$

Final Value of M

Therefore, the value of the expression M is 0 for an acute angle A.