If $tan B =\frac{5}{3}$, what is the value of $\frac{cosec B + sin B}{cos B-sec B}$ ?

$-\frac{177}{125}$

tanB = \(\frac{5}{4}\) = \(\frac{P}{B}\)

Using pythagoras theorem ,

P² + B² = H²

5² + 3² = H²

H² = 34

H = √34

Now ,

$\frac{cosec B + sin B}{cos B-sec B}$

= ( \(\frac{√34 }{5}\) + \(\frac{5 }{ √34}\) ) × ( \(\frac{3 }{√34}\) - \(\frac{√34 }{ 3}\) )

= ( \(\frac{34 +25 }{5√34}\)  × ( \(\frac{9 - 34 }{3√34}\)

= ( \(\frac{59 }{5\)  × ( \(\frac{-25 }{3}\) 

=  - \(\frac{177 }{125\)