Question

If $tan\theta =\frac{5}{12}$, find $\frac{(1+cos\theta )\times (1-sin\theta )}{(1+sin\theta )\times (1-cos\theta )}$.

• A. $\frac{100}{9}$
• B. $\frac{91}{18}$
• C. $\frac{84}{27}$
• D. $\frac{12}{5}$

Answer 1

The correct option is A $\frac{100}{9}$


Consider $\Delta ABC$, right angled at B.

Given: $tan\theta =\frac{5}{12}$

$\Rightarrow$ $tan\theta =\frac{AB}{BC}=\frac{5}{12}$

$\Rightarrow$ If AB = 5k, then BC = 12k, where k is a positive number.

Applying Pythagoras theorem, we have

$A{B}^2+B{C}^2=A{C}^2$

${AC}^2=(12k{)}^2+(5k{)}^2$

${AC}^2=169k^2$

$AC=13k$

$sin\theta =\frac{AB}{AC}=\frac{5k}{13k}=\frac{5}{13}$

$cos\theta =\frac{BC}{AC}=\frac{12k}{13k}=\frac{12}{13}$

$\Rightarrow \frac{(1+cos\theta )\times (1-sin\theta )}{(1+sin\theta )\times (1-cos\theta )}=\frac{(1+\frac{12}{13})\times (1-\frac{5}{13})}{(1+\frac{5}{13})\times (1-\frac{12}{13})}$

$=\frac{\frac{25}{13}\times \frac{8}{13}}{\frac{18}{13}\times \frac{1}{13}}$

$=\frac{200}{18}=\frac{100}{9}$