Question

In ∆ABC, ∠B = 90°, BC =5 cm, 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,
$A{C}^2=A{B}^2+B{C}^2$
$(k+1{)}^2=(k{)}^2+(5{)}^2$
$2k+24$
$k=12$
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
  $\frac{1+SinC}{1+CosC}$ = $\frac{1+\frac{12}{13}}{1+\frac{5}{13}}$
  =$\frac{\frac{13+12}{13}}{\frac{13+5}{13}}$
  =$\frac{25}{18}$