Question

If $sin\theta =\frac{5}{13},$ then the value of $se{c}^2\theta +ta{n}^2\theta$ is

• A.
  1
• B. $\frac{97}{72}$
• C.
  
  $\frac{119}{72}$
• D.
  $\frac{11}{8}$

Answer 1

The correct option is B $\frac{97}{72}$

$sin\theta =\frac{5}{13}$
In a right triangle,

$sin\theta =\frac{Perpendicular\left(P\right)}{Hypotenuse\left(H\right)}=\frac{5}{13}$

Let P=5k and H=13k (where k is any positive integer)



By pythagoras therem,

$H^2=P^2+B^2\Rightarrow (13K{)}^2=(5K{)}^2+B^2\Rightarrow B^2=169k^2-25k^2=144k^2\Rightarrow B=12K\therefore cos\theta =\frac{B}{H}=\frac{12k}{13k}=\frac{12}{13}\Rightarrow sec\theta =\frac{1}{cos\theta }=\frac{13}{12}andtan\theta =\frac{P}{B}=\frac{5k}{12k}=\frac{5}{12}\therefore se{c}^2\theta +ta{n}^2\theta ={\left(\frac{13}{12}\right)}^2+{\left(\frac{5}{12}\right)}^2=\frac{169+25}{144}=\frac{194}{144}=\frac{97}{72}$

Hence, the correct answer is option b.