Question

If 5 secθ – 12 cosecθ = 0, find the values of secθ, cosθ and sinθ.

Answer 1

$$\begin{gathered} 5\sec \theta -12\mathrm{cosec}\theta =0 \\ \Rightarrow 5\sec \theta =12\mathrm{cosec}\theta \\ \Rightarrow \frac{5}{\cos \theta }=\frac{12}{\sin \theta } \\ \Rightarrow \frac{\sin \theta }{\cos \theta }=\frac{12}{5} \\ \Rightarrow \tan \theta =\frac{12}{5} \end{gathered}$$
We have,

$$\begin{gathered} {\sec }^2\theta =1+{\tan }^2\theta \\ \Rightarrow {\sec }^2\theta =1+{\left(\frac{12}{5}\right)}^2 \\ \Rightarrow {\sec }^2\theta =1+\frac{144}{25}=\frac{169}{25} \\ \Rightarrow \sec \theta =\sqrt{\frac{169}{25}}=\frac{13}{5} \end{gathered}$$
Now,

$$\begin{gathered} \cos \theta =\frac{1}{\sec \theta } \\ \Rightarrow \cos \theta =\frac{1}{\left({\displaystyle \frac{13}{5}}\right)} \\ \Rightarrow \cos \theta =\frac{5}{13} \end{gathered}$$
Also,

$$\begin{gathered} \frac{\sin \theta }{\cos \theta }=\tan \theta \\ \Rightarrow \sin \theta =\tan \theta \times \cos \theta \\ \Rightarrow \sin \theta =\frac{12}{5}\times \frac{5}{13}=\frac{12}{13} \end{gathered}$$
Thus, the values of secθ, cosθ and sinθ are $\frac{13}{5}$, $\frac{5}{13}$ and $\frac{12}{13}$, respectively.