Question

In $△ABC$, $\angle B={90}^0,BC=5cm$ and $AC-AB=1$. Evaluate: $\frac{1+sinC}{1+cosC}$

Answer 1

Step 1:

• Here we need the values of sin

• C and cos C and to calculate them we need the values of sides AB and AC. So first we need to find AB and AC.

• It is given that AC - AB = 1.

• Let AB = k and AC = k + 1

Using Pythagoras theorem,

$A{C}^2=A{B}^2+B{C}^2$
$(k+1{)}^2=(k{)}^2+(5{)}^2$
$k^2+1+2k=k^2+25$
$2k=24$
$k=12$

Therefore, AC = 13 cm and AB = 12 cm

Step 2: Next we need to calculate the values of sin C and cos C.
$sinC=\frac{AB}{AC}=\frac{12}{13}$
and,
$cosC=\frac{BC}{AC}=\frac{5}{13}$

• Step 3: For the final step, we need to substitute the corresponding values in the given expression.

We need to find the value of

$\frac{1+sinC}{1+cosC}$

On substituting the values of sin C and cos C, we get,

$\frac{1+sinC}{1+cosC}=\frac{1+\frac{12}{13}}{1+\frac{5}{13}}$

$=\frac{\frac{13+12}{13}}{\frac{13+5}{13}}$

$=\frac{25}{18}$