If 8cotθ = 7 then the value of $\frac{1+sin\theta }{cos\theta }$ is:

The correct option is:

$\frac{\sqrt{113}+8}{7}$

Step-by-step Derivation:

1. We are given the following trigonometric relation:

$8cot\theta =7$

Dividing both sides by 8, we get:

$cot\theta =\frac{7}{8}$

2. We know that the cotangent of an angle in a right-angled triangle is the ratio of the base to the perpendicular side:

$cot\theta =\frac{\text{Base}}{\text{Perpendicular}}=\frac{7}{8}$

Therefore, we can assume the base ($b$) is 7 and the perpendicular ($p$) is 8.

3. Next, we find the hypotenuse ($h$) of the right-angled triangle using the Pythagorean theorem:

$h^2=p^2+b^2$

Substitute the values of $p$ and $b$:

$h^2=8^2+7^2$

$h^2=64+49$

$h^2=113$

Taking the square root of both sides:

$h=\sqrt{113}$

4. Now, we determine the values of sine and cosine for the angle $\theta$:

$sin\theta =\frac{\text{Perpendicular}}{\text{Hypotenuse}}=\frac{8}{\sqrt{113}}$

$cos\theta =\frac{\text{Base}}{\text{Hypotenuse}}=\frac{7}{\sqrt{113}}$

5. Now, we evaluate the expression:

$\frac{1+sin\theta }{cos\theta }$

Substitute the values of $sin\theta$ and $cos\theta$ into the expression:

$\frac{1+\frac{8}{\sqrt{113}}}{\frac{7}{\sqrt{113}}}$

Simplify the numerator by finding a common denominator:

$1+\frac{8}{\sqrt{113}}=\frac{\sqrt{113}+8}{\sqrt{113}}$

Now, divide this numerator by the denominator:

$\frac{\frac{\sqrt{113}+8}{\sqrt{113}}}{\frac{7}{\sqrt{113}}}=\frac{\sqrt{113}+8}{\sqrt{113}}\times \frac{\sqrt{113}}{7}$

Cancel out $\sqrt{113}$ from both the numerator and the denominator:

$\frac{\sqrt{113}+8}{7}$