If `cos37^(@)=(a)/(b)` then , what is the valie of cosec `37^(@)-cos53^(@)` ?

Options

A. `(b^(2)-a^(2))/(ab)`
B. `(a^(2))/(bsqrt(a^(2)+b^(2)))`
C. `(bsqrt(a^(2)+b^(2)))/(a)`
D. `(a^(2))/(bsqrt(b^(2)-a^(2)))`

Answer

To solve the problem, we need to find the value of \( \csc 37^\circ - \cos 53^\circ \) given that \( \cos 37^\circ = \frac{a}{b} \).

### Step-by-Step Solution:

1. **Understand the relationship between angles**:
We know that \( \cos 53^\circ \) can be expressed in terms of \( \sin 37^\circ \) because \( 53^\circ = 90^\circ - 37^\circ \). Therefore, we have:
\[ \cos 53^\circ = \sin 37^\circ \]

2. **Find \( \sin 37^\circ \)**:
Using the Pythagorean identity, we can express \( \sin 37^\circ \) in terms of \( \cos 37^\circ \):
\[ \sin^2 37^\circ + \cos^2 37^\circ = 1 \]
Substituting \( \cos 37^\circ = \frac{a}{b} \):
\[ \sin^2 37^\circ + \left(\frac{a}{b}\right)^2 = 1 \]
Rearranging gives:
\[ \sin^2 37^\circ = 1 - \left(\frac{a}{b}\right)^2 = \frac{b^2 - a^2}{b^2} \]
Taking the square root:
\[ \sin 37^\circ = \frac{\sqrt{b^2 - a^2}}{b} \]

3. **Calculate \( \csc 37^\circ \)**:
The cosecant function is the reciprocal of the sine function:
\[ \csc 37^\circ = \frac{1}{\sin 37^\circ} = \frac{b}{\sqrt{b^2 - a^2}} \]

4. **Substitute into the expression**:
Now we can substitute \( \csc 37^\circ \) and \( \cos 53^\circ \) into the expression we need to evaluate:
\[ \csc 37^\circ - \cos 53^\circ = \frac{b}{\sqrt{b^2 - a^2}} - \sin 37^\circ \]
Since \( \sin 37^\circ = \frac{\sqrt{b^2 - a^2}}{b} \), we can rewrite the expression:
\[ = \frac{b}{\sqrt{b^2 - a^2}} - \frac{\sqrt{b^2 - a^2}}{b} \]

5. **Combine the fractions**:
To combine these two fractions, we need a common denominator:
\[ = \frac{b^2 - (b^2 - a^2)}{b \sqrt{b^2 - a^2}} = \frac{a^2}{b \sqrt{b^2 - a^2}} \]

### Final Result:
Thus, the value of \( \csc 37^\circ - \cos 53^\circ \) is:
\[ \frac{a^2}{b \sqrt{b^2 - a^2}} \]