10th Class Mathematics Introduction to Trigonometry Question Bank Case Based (MCQs) - Introduction to Trigonometry

The value of \[\tan R+\frac{3}{\sec P}-1\]is:

A) \[\frac{253}{136}\]

B) \[\frac{357}{136}\]

C) \[\frac{479}{136}\]

D) \[1\]

Correct Answer: C

Solution :

From part (1) and (3),

\[\cos R=\frac{8}{17}\] and \[\sin R=\frac{15}{17}\]

\[\Rightarrow \,\,\,\,\,\,\,\,\,\,\tan R=\frac{\sin R}{\cos R}=\frac{15/17}{8/17}=\frac{15}{8}\]

From part (3), \[\cos P=\frac{15}{17}\,\,\,\,\,\Rightarrow \,\,\,\,\,\sec P=\frac{17}{15}\]

\[\therefore \,\,\,\,\,\,\,\,\,\,\,\tan R+\frac{3}{\sec P}-1=\frac{15}{8}+\frac{3}{(17/15)}-1\]

\[=\frac{15}{8}+\frac{45}{17}-1=\frac{255+360-136}{136}=\frac{479}{136}\]

So, option [c] is correct.