For 0<θ<π/2, the solution (s) of ∑ₘ₌₁⁶cosec(θ+((m-1)π)/4)cosec(θ+mπ/4)=4√2 is (are)
0 < θ < π 2 0<\theta<\dfrac{\pi}{2} 0 < θ < 2 π ∑ m = 1 6 cosec ( θ + ( m − 1 ) π 4 ) cosec ( θ + m π 4 ) = 4 2 \displaystyle\sum_{m=1}^{6}\operatorname{cosec}\left(\theta+\dfrac{(m-1)\pi}{4}\right)\operatorname{cosec}\left(\theta+\dfrac{m\pi}{4}\right)=4\sqrt{2} m = 1 ∑ 6 cosec ( θ + 4 ( m − 1 ) π ) cosec ( θ + 4 mπ ) = 4 2 To find: the value(s) of θ \text{To find: the value(s) of } \theta To find: the value(s) of θ ( 1 ) π 4 ( 2 ) π 6 ( 3 ) π 12 ( 4 ) 5 π 12 (\text{1})\ \dfrac{\pi}{4}\quad(\text{2})\ \dfrac{\pi}{6}\quad(\text{3})\ \dfrac{\pi}{12}\quad(\text{4})\ \dfrac{5\pi}{12} ( 1 ) 4 π ( 2 ) 6 π ( 3 ) 12 π ( 4 ) 12 5 π Plan: (i) cosec A cosec B = cot A − cot B sin ( B − A ) \text{Plan: (i) } \operatorname{cosec}A\,\operatorname{cosec}B=\dfrac{\cot A-\cot B}{\sin(B-A)} Plan: (i) cosec A cosec B = sin ( B − A ) cot A − cot B (ii) apply to all 6 terms ( B − A = π / 4 ) \text{(ii) apply to all 6 terms }(B-A=\pi/4) (ii) apply to all 6 terms ( B − A = π /4 ) (iii) telescope, then solve for θ \text{(iii) telescope, then solve for }\theta (iii) telescope, then solve for θ cot A − cot B = cos A sin A − cos B sin B \cot A-\cot B=\dfrac{\cos A}{\sin A}-\dfrac{\cos B}{\sin B} cot A − cot B = sin A cos A − sin B cos B = cos A sin B − cos B sin A sin A sin B =\dfrac{\cos A\sin B-\cos B\sin A}{\sin A\sin B} = sin A sin B cos A sin B − cos B sin A = sin ( B − A ) sin A sin B =\dfrac{\sin(B-A)}{\sin A\sin B} = sin A sin B sin ( B − A ) ⇒ cosec A cosec B = cot A − cot B sin ( B − A ) \Rightarrow\ \operatorname{cosec}A\,\operatorname{cosec}B=\dfrac{\cot A-\cot B}{\sin(B-A)} ⇒ cosec A cosec B = sin ( B − A ) cot A − cot B A = θ + ( m − 1 ) π 4 , B = θ + m π 4 ⇒ B − A = π 4 A=\theta+\dfrac{(m-1)\pi}{4},\quad B=\theta+\dfrac{m\pi}{4}\ \Rightarrow\ B-A=\dfrac{\pi}{4} A = θ + 4 ( m − 1 ) π , B = θ + 4 mπ ⇒ B − A = 4 π sin π 4 = 1 2 ⇒ cosec A cosec B = 2 ( cot A − cot B ) \sin\dfrac{\pi}{4}=\dfrac{1}{\sqrt{2}}\ \Rightarrow\ \operatorname{cosec}A\,\operatorname{cosec}B=\sqrt{2}(\cot A-\cot B) sin 4 π = 2 1 ⇒ cosec A cosec B = 2 ( cot A − cot B ) S = ∑ m = 1 6 cosec A m cosec B m S=\displaystyle\sum_{m=1}^{6}\operatorname{cosec}A_m\,\operatorname{cosec}B_m S = m = 1 ∑ 6 cosec A m cosec B m S = 2 ∑ m = 1 6 ( cot A m − cot B m ) S=\sqrt{2}\sum_{m=1}^{6}\left(\cot A_m-\cot B_m\right) S = 2 ∑ m = 1 6 ( cot A m − cot B m ) S = 2 [ ( cot θ − cot ( θ + π 4 ) ) + ( cot ( θ + π 4 ) − cot ( θ + 2 π 4 ) ) + ⋯ ] S=\sqrt{2}\Big[(\cot\theta-\cot(\theta+\tfrac{\pi}{4}))+(\cot(\theta+\tfrac{\pi}{4})-\cot(\theta+\tfrac{2\pi}{4}))+\cdots\Big] S = 2 [ ( cot θ − cot ( θ + 4 π )) + ( cot ( θ + 4 π ) − cot ( θ + 4 2 π )) + ⋯ ] consecutive terms cancel pairwise \text{consecutive terms cancel pairwise} consecutive terms cancel pairwise S = 2 [ cot θ − cot ( θ + 6 π 4 ) ] S=\sqrt{2}\Big[\cot\theta-\cot\left(\theta+\dfrac{6\pi}{4}\right)\Big] S = 2 [ cot θ − cot ( θ + 4 6 π ) ] 6 π 4 = 3 π 2 = π + π 2 \dfrac{6\pi}{4}=\dfrac{3\pi}{2}=\pi+\dfrac{\pi}{2} 4 6 π = 2 3 π = π + 2 π cot ( θ + 3 π 2 ) = cot ( θ + π 2 ) = − tan θ \cot\left(\theta+\dfrac{3\pi}{2}\right)=\cot\left(\theta+\dfrac{\pi}{2}\right)=-\tan\theta cot ( θ + 2 3 π ) = cot ( θ + 2 π ) = − tan θ S = 2 [ cot θ − ( − tan θ ) ] = 2 ( cot θ + tan θ ) S=\sqrt{2}\big[\cot\theta-(-\tan\theta)\big]=\sqrt{2}(\cot\theta+\tan\theta) S = 2 [ cot θ − ( − tan θ ) ] = 2 ( cot θ + tan θ ) cot θ + tan θ = cos θ sin θ + sin θ cos θ \cot\theta+\tan\theta=\dfrac{\cos\theta}{\sin\theta}+\dfrac{\sin\theta}{\cos\theta} cot θ + tan θ = sin θ cos θ + cos θ sin θ = cos 2 θ + sin 2 θ sin θ cos θ = 1 sin θ cos θ =\dfrac{\cos^2\theta+\sin^2\theta}{\sin\theta\cos\theta}=\dfrac{1}{\sin\theta\cos\theta} = sin θ cos θ cos 2 θ + sin 2 θ = sin θ cos θ 1 sin θ cos θ = 1 2 sin 2 θ ⇒ cot θ + tan θ = 2 sin 2 θ \sin\theta\cos\theta=\dfrac{1}{2}\sin 2\theta\ \Rightarrow\ \cot\theta+\tan\theta=\dfrac{2}{\sin 2\theta} sin θ cos θ = 2 1 sin 2 θ ⇒ cot θ + tan θ = sin 2 θ 2 S = 2 ⋅ 2 sin 2 θ = 2 2 sin 2 θ S=\sqrt{2}\cdot\dfrac{2}{\sin 2\theta}=\dfrac{2\sqrt{2}}{\sin 2\theta} S = 2 ⋅ sin 2 θ 2 = sin 2 θ 2 2 2 2 sin 2 θ = 4 2 ⇒ sin 2 θ = 1 2 \dfrac{2\sqrt{2}}{\sin 2\theta}=4\sqrt{2}\ \Rightarrow\ \sin 2\theta=\dfrac{1}{2} sin 2 θ 2 2 = 4 2 ⇒ sin 2 θ = 2 1 0 < θ < π 2 ⇒ 0 < 2 θ < π 0<\theta<\dfrac{\pi}{2}\ \Rightarrow\ 0<2\theta<\pi 0 < θ < 2 π ⇒ 0 < 2 θ < π sin 2 θ = 1 2 , sin > 0 on ( 0 , π ) ⇒ two solutions \sin 2\theta=\dfrac{1}{2},\ \sin>0 \text{ on }(0,\pi)\ \Rightarrow\ \text{two solutions} sin 2 θ = 2 1 , sin > 0 on ( 0 , π ) ⇒ two solutions 2 θ = π 6 or 2 θ = 5 π 6 2\theta=\dfrac{\pi}{6}\ \text{ or }\ 2\theta=\dfrac{5\pi}{6} 2 θ = 6 π or 2 θ = 6 5 π θ = π 12 or θ = 5 π 12 \theta=\dfrac{\pi}{12}\ \text{ or }\ \theta=\dfrac{5\pi}{12} θ = 12 π or θ = 12 5 π θ = π 12 and θ = 5 π 12 \boxed{\theta=\dfrac{\pi}{12}\ \text{ and }\ \theta=\dfrac{5\pi}{12}} θ = 12 π and θ = 12 5 π options (3) and (4) \text{options (3) and (4)} options (3) and (4)
