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5. Given \sec \theta =\frac{13}{12}, calculate all other trigonometric ratios.

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We have,
\sec \theta =\frac{13}{12},

 It means Hypotenuse of the triangle is 13 units and the base is 12 units.
Let ABC is a right-angled triangle in which \angleB is 90 and AB is the base, BC is perpendicular height and AC is the hypotenuse.


By using Pythagoras theorem,

BC = \sqrt{169-144}=\sqrt{25}
BC = 5 unit

Therefore,
\sin \theta = \frac{BC}{AC}=\frac{5}{13} 
\cos \theta = \frac{BA}{AC}=\frac{12}{13}

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

\cot \theta = \frac{BA}{BC}=\frac{12}{5}

\sec \theta = \frac{AC}{AB}=\frac{13}{12}

\csc \theta = \frac{AC}{BC}=\frac{13}{5}

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manish

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