Question
If $\sec \theta=\frac{13}{12}$ find the values of other trigonometric ratios.
✓
Answer
We know that,
$\sec ^2 \theta=1+\tan ^2 \theta$
$\Rightarrow \tan ^2 \theta=\sec ^2 \theta-1$
$\Rightarrow \tan ^2 \theta=\left(\frac{13}{12}\right)^2-1$
$\Rightarrow \tan ^2 \theta=\frac{169}{144}-1=\frac{25}{144}$
$\Rightarrow \tan \theta=\frac{5}{12} \ldots$
Also,
$\cos \theta=\frac{1}{\sec \theta}$
$\Rightarrow \cos \theta=\frac{12}{13} \ldots$
Now, using
$\tan \theta=\frac{\sin \theta}{\cos \theta}$
$\Rightarrow \frac{5}{12}=\frac{\sin \theta}{\frac{12}{13}}$
$\Rightarrow \sin \theta=\frac{5}{12} \times \frac{12}{13}$
$\Rightarrow \sin \theta=\frac{5}{13} \ldots[3]$
Also,
$\operatorname{cosec} \theta=\frac{1}{\sin \theta}=\frac{13}{5} \ldots$
$\cot \theta=\frac{1}{\tan \theta}=\frac{12}{5}$
$\sec ^2 \theta=1+\tan ^2 \theta$
$\Rightarrow \tan ^2 \theta=\sec ^2 \theta-1$
$\Rightarrow \tan ^2 \theta=\left(\frac{13}{12}\right)^2-1$
$\Rightarrow \tan ^2 \theta=\frac{169}{144}-1=\frac{25}{144}$
$\Rightarrow \tan \theta=\frac{5}{12} \ldots$
Also,
$\cos \theta=\frac{1}{\sec \theta}$
$\Rightarrow \cos \theta=\frac{12}{13} \ldots$
Now, using
$\tan \theta=\frac{\sin \theta}{\cos \theta}$
$\Rightarrow \frac{5}{12}=\frac{\sin \theta}{\frac{12}{13}}$
$\Rightarrow \sin \theta=\frac{5}{12} \times \frac{12}{13}$
$\Rightarrow \sin \theta=\frac{5}{13} \ldots[3]$
Also,
$\operatorname{cosec} \theta=\frac{1}{\sin \theta}=\frac{13}{5} \ldots$
$\cot \theta=\frac{1}{\tan \theta}=\frac{12}{5}$
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