If $\frac{\sec A+\tan A}{\sec A-\tan A}=2^{\frac{51}{79}},$ then the value of sin A is equal to:

The correct answer is 65/144.

Given:

$\frac{\sec A+\tan A}{\sec A-\tan A}=2\frac{51}{79}$

First, convert the mixed fraction on the right-hand side into an improper fraction:

$2\frac{51}{79}=\frac{(2\times 79)+51}{79}=\frac{158+51}{79}=\frac{209}{79}$

So, the given equation becomes:

$\frac{\sec A+\tan A}{\sec A-\tan A}=\frac{209}{79}$

Apply the rule of Componendo and Dividendo (if $\frac{a}{b}=\frac{c}{d}$, then $\frac{a+b}{a-b}=\frac{c+d}{c-\mathrm{d}}$):

$\frac{(\sec A+\tan A)+(\sec A-\tan A)}{(\sec A+\tan A)-(\sec A-\tan A)}=\frac{209+79}{209-79}$

Simplify the numerator and denominator on both sides:

$\frac{2\sec A}{2\tan A}=\frac{288}{130}$

$\frac{\sec A}{\tan A}=\frac{144}{65}$

Now, express $\sec A$ and $\tan A$ in terms of $\sin A$ and $\cos A$:

$\frac{\frac{1}{\cos A}}{\frac{\sin A}{\cos A}}=\frac{144}{65}$

$\frac{1}{\sin A}=\frac{144}{65}$

Taking the reciprocal on both sides gives:

$\sin A=\frac{65}{144}$

Therefore, the value of sin A is equal to 65/144.