question_answer

In \[\Delta PQR\], \[\angle Q=90{}^\circ \]. If PQ = 10 cm and PR =15 cm. Then, the value of \[{{\tan }^{2}}P+{{\sec }^{2}}P+1\]is

A) \[\frac{5}{2}\]

B) \[\frac{3}{17}\]

C) \[\frac{9}{2}\]

D) \[\frac{4}{5}\]

Correct Answer: C

Solution :

Given, \[PQ=10\,cm\], PR = 15 cm

Apply Pythagoras theorem,

\[P{{Q}^{2}}\text{ }+\text{ }Q{{R}^{2}}\text{ }=\text{ }P{{R}^{2}}\]

\[\Rightarrow \,\,\,\,\,\,{{10}^{2}}+Q{{R}^{2}}={{15}^{2}}\]

\[\Rightarrow \,\,\,\,\,\,QR=\sqrt{225-100}\]

\[=\sqrt{125}\]

\[\therefore \,\,{{\tan }^{2}}+{{\sec }^{2}}p+1\]

\[={{\left( \frac{QR}{PQ} \right)}^{2}}+{{\left( \frac{PR}{PQ} \right)}^{2}}+1\]

\[=\frac{125}{100}+\frac{225}{100}+1\]

\[=\frac{125+225+100}{100}=\frac{450}{100}=\frac{9}{2}\]