JEE Advanced 2009 Paper 2
Exam JEE Advanced
Year 2009
Paper Paper 2
Subject Mathematics
Chapter Trigonometric Equations JEE Advanced 2009 · Paper 2 · Mathematics · Trigonometric Equations
For 0 < θ < (π)/(2) , the solution(s) of ∑_ m=1 ^6 cosec[θ+((m-1)...
For
0 < θ < π 2 0 < \theta < \frac{\pi}{2} 0 < θ < 2 π , the solution(s)
o f ∑ m = 1 6 cosec [ θ + ( m − 1 ) π 4 ] of \sum_{m=1}^6 \operatorname{cosec}\left[\theta+\frac{(m-1) \pi}{4}\right] o f ∑ m = 1 6 cosec [ θ + 4 ( m − 1 ) π ] cosec ( θ + m π 4 ) = 4 2 \operatorname{cosec}\left(\theta+\frac{m \pi}{4}\right)=4 \sqrt{2} cosec ( θ + 4 mπ ) = 4 2 is/are
Options A π 4 \frac{\pi}{4} 4 π B π 6 \frac{\pi}{6} 6 π C π 12 \frac{\pi}{12} 12 π D 5 π 12 \frac{5 \pi}{12} 12 5 π Correct answer C. π 12 \frac{\pi}{12} 12 π
Step-by-step solution
For
0 < θ < π 2 0 < \theta < \frac{\pi}{2} 0 < θ < 2 π ,
∑ m = 1 6 cosec [ θ + ( m − 1 ) π 4 ] cosec ( θ + m π 4 ) = 4 2 ⇒ ∑ m = 1 6 sin [ θ + m π 4 − ( θ + ( m − 1 ) π 4 ) ] sin π 4 { sin ( θ + ( m − 1 ) π 4 ) sin ( θ + m π 4 ) } = 4 2 \begin{array}{r}
\sum_{m=1}^6 \operatorname{cosec}\left[\theta+\frac{(m-1) \pi}{4}\right]\operatorname{cosec}\left(\theta+\frac{m \pi}{4}\right)\\
=4 \sqrt{2}\\
\Rightarrow \sum_{m=1}^6 \frac{\sin \left[\theta+\frac{m \pi}{4}-\left(\theta+\frac{(m-1) \pi}{4}\right)\right]}{\sin \frac{\pi}{4}\left\{\sin \left(\theta+\frac{(m-1) \pi}{4}\right) \sin \left(\theta+\frac{m \pi}{4}\right)\right\}}\\
=4 \sqrt{2}
\end{array} ∑ m = 1 6 cosec [ θ + 4 ( m − 1 ) π ] cosec ( θ + 4 mπ ) = 4 2 ⇒ ∑ m = 1 6 s i n 4 π { s i n ( θ + 4 ( m − 1 ) π ) s i n ( θ + 4 mπ ) } s i n [ θ + 4 mπ − ( θ + 4 ( m − 1 ) π ) ] = 4 2 ⇒ ∑ m = 1 6 cot ( θ + ( m − 1 ) π 4 ) − cot ( θ + m π 4 ) 1 / 2 = 4 2 \begin{array}{r}
\Rightarrow \sum_{m=1}^6 \frac{\cot \left(\theta+\frac{(m-1) \pi}{4}\right)-\cot \left(\theta+\frac{m \pi}{4}\right)}{1 / 2}\\
=4 \sqrt{2}
\end{array} ⇒ ∑ m = 1 6 1/2 c o t ( θ + 4 ( m − 1 ) π ) − c o t ( θ + 4 mπ ) = 4 2 ⇒ ∑ m = 1 6 [ cot ( θ + ( m − 1 ) π 4 ) − cot ( θ + m π 4 ) ] = 4 ⇒ cot ( θ ) − cot ( θ + π 4 ) + cot ( θ + π 4 ) − cot ( θ + 2 π 4 ) + … + cot ( θ + 5 π 4 ) − cot ( θ + 6 π 4 ) = 4 ⇒ cot θ − cot ( 3 π 2 + θ ) = 4 ⇒ cot θ + tan θ = 4 ⇒ tan 2 θ − 4 tan θ + 1 = 0 ⇒ ( tan θ − 2 ) 2 − 3 = 0 ⇒ ( tan \ttheta − 2 + 3 ) ( tan \ttheta − 2 − 3 ) = 0 ⇒ tan \ttheta = 2 − 3 or tan \t t h e t a = 2 + 3 ⇒ θ = π 12 ; θ = 5 π 12 ⇒ θ ∈ ( 0 , π 2 ) \begin{aligned}& \Rightarrow \quad \sum_{m=1}^6\left[\cot \left(\theta+\frac{(m-1) \pi}{4}\right)\right.\\
& \left.-\cot \left(\theta+\frac{m \pi}{4}\right)\right]=4 \\
& \Rightarrow \cot (\theta)-\cot \left(\theta+\frac{\pi}{4}\right)+\cot \left(\theta+\frac{\pi}{4}\right)\\
& -\cot \left(\theta+\frac{2 \pi}{4}\right)+\ldots+\cot \left(\theta+\frac{5 \pi}{4}\right)\\
& -\cot \left(\theta+\frac{6 \pi}{4}\right)=4 \\
& \Rightarrow \quad \cot \theta-\cot \left(\frac{3 \pi}{2}+\theta\right)=4 \\
& \Rightarrow \quad \cot \theta+\tan \theta=4 \\
& \Rightarrow \quad \tan ^2 \theta-4 \tan \theta+1=0 \\
& \Rightarrow \quad(\tan \theta-2)^2-3=0 \\
& \Rightarrow(\tan \ttheta-2+\sqrt{3})(\tan \ttheta-2-\sqrt{3})=0 \\
& \Rightarrow \quad \tan \ttheta=2-\sqrt{3} \text{ or }\\
& \tan \t theta=2+\sqrt{3}\\
& \Rightarrow \theta=\frac{\pi}{12}; \theta=\frac{5 \pi}{12} \Rightarrow \theta \in \left(0, \frac{\pi}{2}\right)\\
&
\end{aligned} ⇒ m = 1 ∑ 6 [ cot ( θ + 4 ( m − 1 ) π ) − cot ( θ + 4 mπ ) ] = 4 ⇒ cot ( θ ) − cot ( θ + 4 π ) + cot ( θ + 4 π ) − cot ( θ + 4 2 π ) + … + cot ( θ + 4 5 π ) − cot ( θ + 4 6 π ) = 4 ⇒ cot θ − cot ( 2 3 π + θ ) = 4 ⇒ cot θ + tan θ = 4 ⇒ tan 2 θ − 4 tan θ + 1 = 0 ⇒ ( tan θ − 2 ) 2 − 3 = 0 ⇒ ( tan \ttheta − 2 + 3 ) ( tan \ttheta − 2 − 3 ) = 0 ⇒ tan \ttheta = 2 − 3 or tan \t t h e t a = 2 + 3 ⇒ θ = 12 π ; θ = 12 5 π ⇒ θ ∈ ( 0 , 2 π )
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