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  • The number of solutions of <img alt="\tan (5 \pi \cos \alpha)=\cot (5 \pi \sin \alpha)\: f\! or\: \alpha \: in\: (0,2 \pi) \; \; is" src="https://entrancecorner.oncodecogs.com/gif.latex?%

The number of solutions of

\tan (5 \pi \cos \alpha)=\cot (5 \pi \sin \alpha)\: f\! or\: \alpha \: in\: (0,2 \pi) \; \; is is

Option: 1

7


Option: 2

14


Option: 3

21


Option: 4

28


Answers (1)

best_answer

\begin{aligned} & \tan (5 \pi \cos \alpha)=\cot (5 \pi \sin \alpha) \\ & \Rightarrow \quad \cot (5 \pi \sin \alpha)-\tan (5 \pi \cos \alpha)=0 \\ & \Rightarrow \quad \frac{\cos (5 \pi \sin \alpha)}{\sin (5 \pi \sin \alpha)}-\frac{\sin (5 \pi \cos \alpha)}{\cos (5 \pi \cos \alpha)}=0 \\ & \Rightarrow \quad \cos \{5 \pi(\sin \alpha+\cos \alpha)\}=0 \\ & \Rightarrow5 \pi(\sin \alpha+\cos \alpha)=n \pi+\frac{\pi}{2}, n \in I \\ & \end{aligned}

\begin{aligned} & \Rightarrow \quad \sin \alpha+\cos \alpha=\frac{(2 n+1)}{10} \\ & \Rightarrow \quad \frac{1}{\sqrt{2}} \sin \alpha+\frac{1}{\sqrt{2}} \cos \alpha=\frac{(2 n+1)}{10 \sqrt{2}} \end{aligned}

\begin{aligned} & \therefore \quad \sin \left(\frac{\pi}{4}+\alpha\right)=\frac{(2 n+1)}{10 \sqrt{2}} ------I\\ & \therefore \quad \frac{\pi}{4}+\alpha=n \pi+(-1)^n \sin ^{-1}\left(\frac{2 n+1}{10 \sqrt{2}}\right) \\ & \Rightarrow \alpha=n \pi-\frac{\pi}{4}+(-1)^n \sin ^{-1}\left(\frac{2 n+1}{10 \sqrt{2}}\right), n \in I--------II \\ & \end{aligned}

From eqution (I)

\begin{array}{cc} & -1 \leq \sin \left(\frac{\pi}{4}+\alpha\right) \leq 1 \\ \\\Rightarrow & -1 \leq \frac{(2 n+1)}{10 \sqrt{2}} \leq 1 \\ \\\Rightarrow & \frac{-10 \sqrt{2}-1}{2} \leq n \leq \frac{10 \sqrt{2}-1}{2} \end{array}

\begin{array}{cc} \because & n \in I \\ \therefore & n=-7,-6,-5,-4,-3,-2,-1, \\ & 0,1,2,3,4,5,6 \end{array}

Now, from Eq. (ii)

 Number of solutions   = Number of values of n.
                                         =14 .

Posted by

Suraj Bhandari

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