A) \[\frac{9}{4}\]
B) \[\frac{5}{3}\]
C) \[\frac{2}{5}\]
D) \[\frac{31}{15}\]
Correct Answer: D
Solution :
| We know the identity, |
| \[1+{{\tan }^{2}}S={{\sec }^{2}}S\,\,\,\Rightarrow \,\,\,\,\tan S=\sqrt{{{\sec }^{2}}S-1}=\sqrt{{{\left( \frac{\sqrt{34}}{3} \right)}^{2}}-1}\] |
| [From part 1] |
| \[=\sqrt{\frac{34}{9}-1}=\sqrt{\frac{25}{9}}=\frac{5}{3}\] |
| Also, know the identity, \[1+{{\cot }^{2}}R=\cos e{{c}^{2}}R\] |
| \[\Rightarrow \,\,\,\cot R=\sqrt{\cos e{{c}^{2}}R-1}=\sqrt{{{\left( \frac{\sqrt{29}}{5} \right)}^{2}}-1}=\sqrt{\frac{29}{25}-1}=\sqrt{\frac{4}{25}}=\frac{2}{5}\] |
| \[\therefore \,\,\,\,\,\,\,\,\,\tan S+\cot R=\frac{5}{3}+\frac{2}{5}=\frac{25+6}{15}=\frac{31}{15}\] |
| So, option [d] is correct. |
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