A) 2
B) 3
C) \[\frac{1}{3}\]
D) 5
E) \[\frac{1}{5}\]
Correct Answer: C
Solution :
\[\sin t+\cos t=\frac{1}{5}\] \[\Rightarrow \] \[\frac{2\tan \frac{t}{2}}{1{{\tan }^{2}}\frac{t}{2}}+\frac{1-{{\tan }^{2}}\frac{t}{2}}{1+{{\tan }^{2}}\frac{t}{2}}=\frac{1}{5}\] \[\Rightarrow \] \[5\left( 2\tan \frac{t}{2}+1-{{\tan }^{2}}\frac{t}{2} \right)=1+{{\tan }^{2}}\frac{t}{2}\] \[\Rightarrow \] \[10\tan \frac{t}{2}+5-5{{\tan }^{2}}\frac{t}{2}=1+{{\tan }^{2}}\frac{t}{2}\] \[\Rightarrow \] \[6{{\tan }^{2}}\frac{t}{2}-10\tan \frac{t}{2}-4=0\] \[\Rightarrow \] \[6{{\tan }^{2}}\frac{t}{2}-12\tan \frac{t}{2}-2\tan \frac{t}{2}-4=0\] \[\Rightarrow \] \[6\tan \frac{t}{2}\left( \tan \frac{t}{2}-2 \right)-2\left( \tan \frac{t}{2}-2 \right)=0\] \[\Rightarrow \] \[\left( 6\tan \frac{t}{2}-2 \right)\left( \tan \frac{t}{2}-2 \right)=0\] \[\Rightarrow \] \[\tan \frac{t}{2}=\frac{1}{3},2\]for \[0<t<\frac{\pi }{2}\] \[\tan \frac{t}{2}=\frac{1}{3}\]
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