A) tan M
B) tan2M
C) \[{{\tan }^{2}}M\]
D) None of these
Correct Answer: C
Solution :
| \[{{\sec }^{2}}M-1={{\left( \frac{KM}{LM} \right)}^{2}}-1\] |
| \[{{\sec }^{2}}M-1=\frac{36}{27}-1=\frac{36-27}{27}=\frac{9}{27}\] |
| \[{{\sec }^{2}}M-1=\frac{1}{3}\] |
| \[\tan M=\frac{1}{\sqrt{3}}\] |
| \[{{\tan }^{2}}M=\frac{1}{3}\Rightarrow \,{{\sec }^{2}}M-1{{\tan }^{2}}M\] |
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