If tan mθ = tan nθ , then general value of θ will be in

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Trigonometrical Equations

normal

If $\tan m\theta = \tan n\theta $, then the general value of $\theta $ will be in

$A. P.$

$G. P.$

$H. P.$

None of these

Solution

(a) $\tan m\theta = \tan n\theta $$ \Rightarrow $ $m\theta = p\pi + n\theta \Rightarrow \theta = \frac{{p\pi }}{{(m – n)}}$
Hence different values of $\theta $ are in A.P. with $\frac{\pi }{{m – n}}$ as common difference.

Standard 11

Mathematics

The general value of $\theta $ satisfying ${\sin ^2}\theta + \sin \theta = 2$ is

easy

For each positive real number $\lambda$. Let $A_\lambda$ be the set of all natural numbers $n$ such that $|\sin (\sqrt{n+1})-\sin (\sqrt{n})|<\lambda$. Let $A_\lambda^c$ be the complement of $A_\lambda$ in the set of all natural numbers. Then,

normal

(KVPY-2016)

The set of angles btween $0$ & $2\pi $ satisfying the equation $4\, cos^2 \, \theta – 2 \sqrt 2 \, cos \,\theta – 1 = 0$ is

normal

$\sin 6\theta + \sin 4\theta + \sin 2\theta = 0,$ then $\theta = $

medium

If $\mathrm{n}$ is the number of solutions of the equation

$2 \cos x\left(4 \sin \left(\frac{\pi}{4}+x\right) \sin \left(\frac{\pi}{4}-x\right)-1\right)=1, x \in[0, \pi]$

and $S$ is the sum of all these solutions, then the ordered pair $(\mathrm{n}, \mathrm{S})$ is :

hard

(JEE MAIN-2021)

If $\tan m\theta = \tan n\theta $, then the general value of $\theta $ will be in

Solution

Similar Questions

The general value of $\theta $ satisfying ${\sin ^2}\theta + \sin \theta = 2$ is

For each positive real number $\lambda$. Let $A_\lambda$ be the set of all natural numbers $n$ such that $|\sin (\sqrt{n+1})-\sin (\sqrt{n})|<\lambda$. Let $A_\lambda^c$ be the complement of $A_\lambda$ in the set of all natural numbers. Then,

The set of angles btween $0$ & $2\pi $ satisfying the equation $4\, cos^2 \, \theta – 2 \sqrt 2 \, cos \,\theta – 1 = 0$ is

$\sin 6\theta + \sin 4\theta + \sin 2\theta = 0,$ then $\theta = $

If $\mathrm{n}$ is the number of solutions of the equation $2 \cos x\left(4 \sin \left(\frac{\pi}{4}+x\right) \sin \left(\frac{\pi}{4}-x\right)-1\right)=1, x \in[0, \pi]$ and $S$ is the sum of all these solutions, then the ordered pair $(\mathrm{n}, \mathrm{S})$ is :

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If $\mathrm{n}$ is the number of solutions of the equation

$2 \cos x\left(4 \sin \left(\frac{\pi}{4}+x\right) \sin \left(\frac{\pi}{4}-x\right)-1\right)=1, x \in[0, \pi]$

and $S$ is the sum of all these solutions, then the ordered pair $(\mathrm{n}, \mathrm{S})$ is :