A) Only I
B) Only II
C) Both I and II
D) Neither I nor II
Correct Answer: C
Solution :
Given, A \[\frac{\pi }{6}\]and B\[\frac{\pi }{3}\] |
I. LHS, |
\[\sin A+\sin B=\sin \frac{\pi }{6}+\sin \frac{\pi }{3}=\frac{1}{2}+\frac{\sqrt{3}}{2}=\frac{1+\sqrt{3}}{2}\] |
RHS, \[\cos A+\cos B=\cos \frac{\pi }{6}+\cos \frac{\pi }{3}=\frac{\sqrt{3}}{2}+\frac{1}{2}=\frac{\sqrt{3}+1}{2}\]\[\Rightarrow \]\[\sin A+\sin B=\cos A+\cos B\] |
II. LHS, \[\tan A+\tan B=\tan \frac{\pi }{6}+\tan \frac{\pi }{3}\] |
\[=\frac{1}{\sqrt{3}}+\sqrt{3}=\frac{4}{\sqrt{3}}\] |
RHS, \[\cot A+\cot B=\cot \frac{\pi }{6}+\cot \frac{\pi }{3}=\sqrt{3}+\frac{1}{\sqrt{3}}=\frac{4}{\sqrt{3}}\] |
\[\Rightarrow \]\[\tan \,\,A+tan\,\,B=cot\,\,A+cot\,\,B\] |
Hence, both statements are true. |
Shortcut method |
\[A+B=\frac{\pi }{6}+\frac{\pi }{3}=\frac{\pi }{2}\] |
I. \[\sin A+\sin B=\sin \left( \frac{\pi }{2}-B \right)+\sin \left( \frac{\pi }{2}-A \right)\] |
\[=\cos B+\cos A=\cos A+\cos B\] |
II. \[\tan A+\tan B=\tan \left( \frac{\pi }{2}-B \right)+\tan \left( \frac{\pi }{2}-A \right)\] |
\[=\cot B+\cot A=\cot A+\cot B\] |
\[\therefore \]Both statements are true. |
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