1. Tardigrade
  2. Question
  3. Mathematics
  4. When ( sin 9 θ/ cos 27 θ)+( sin 3 θ/ cos 9 θ)+( sin θ/ cos 3 θ)=k ( tan 27 θ- tan θ) is defined, then k=

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. When $\frac{\sin \,9 \theta}{\cos \,27 \theta}+\frac{\sin \,3 \theta}{\cos \,9 \theta}+\frac{\sin \theta}{\cos \,3 \theta}=k$
$(\tan\, 27 \theta-\tan \,\theta)$ is defined, then $k=$

TS EAMCET 2018

Solution:

Given, $\frac{\sin \,9 \,\theta}{\cos \,27 \,\theta}+\frac{\sin \,3 \,\theta}{\cos \,9 \,\theta}+\frac{\sin \,\theta}{\cos \,3\, \theta}=k$
$(\tan \,27 \,\theta-\tan \,\theta)$
$\because LHS =\frac{\sin \,9 \,\theta \cos \,9 \,\theta+\sin \,3 \theta \,\cos \,27\, \theta}{\cos \,27 \,\theta \,\cos \,9 \,\theta}+\frac{\sin\, \theta}{\cos\, 3\, \theta}$
$=\frac{\sin \,18 \,\theta+\sin \,30 \,\theta-\sin \,24 \,\theta}{2 \,\cos \,9 \,\theta \,\cos \,27 \theta}+\frac{\sin \,\theta}{\cos \,3 \,\theta}$
$ [\sin \,21\,\theta+\sin \,5 \,\theta+\sin \,33 \,\theta+\sin\,27 \,\theta-\sin\, 27\,\theta$
$= \frac{-\sin \,21 \,\theta+4 \,\sin \,\theta \cos \,9 \,\theta \,\cos \,27 \,\theta]}{4 \,\cos\, 3 \,\theta \,\cos \,9 \theta\,\cos \,27 \,\theta} $
$= \frac{\sin 15 \theta+\sin 33 \theta+4 \sin\, \theta\, \cos \,9 \,\theta \,\cos \,27 \,\theta}{4 \cos 3 \theta \,\cos \,9 \,\theta \,\cos \,27 \,\theta} $
$= \frac{2 \,\sin \,24 \,\theta \,\cos \,9 \,\theta+4 \,\sin \,\theta \,\cos \,9 \,\theta\, \cos \,27 \,\theta}{4 \,\cos \,3 \,\theta \,\cos \,9 \theta \,\cos \,27 \,\theta} $4
$= \frac{\sin \,24 \,\theta+2 \,\sin \,\theta \,\cos \,27 \,\theta}{2\,\cos \,3 \,\theta\,\cos \,27\,\theta} $
$= \frac{\sin \,(27 \,\theta-3 \,\theta)}{2 \,\cos \,3 \theta\,\cos\, 27 \,\theta}+\frac{\sin \,\theta}{\cos \,3 \,\theta} $
$=\frac{1}{2}\left[\frac{\sin \,27 \,\theta}{\cos \,27 \,\theta}-\frac{\sin \,3 \,\theta}{\cos 3 \theta}\right]+\frac{\sin \,\theta}{\cos \,3 \,\theta}$
$=\frac{1}{2} \frac{\sin \,27\, \theta}{\cos \,27 \,\theta}-\frac{1}{2}\left(\frac{\sin \,3\,\theta-2\, \sin \,\theta}{\cos \,3\, \theta}\right)$
$=\frac{1}{2} \tan \,27 \,\theta-\frac{1}{2} \frac{3 \,\sin \,\theta-4 \,\sin ^{3} \,\theta-2 \,\sin \,\theta}{\cos \,\theta\left(4 \,\cos ^{2}\, \theta-3\right)}$
$=\frac{1}{2}\left[\tan\, 27 \,\theta-\frac{\sin \,\theta\left(1-4 \,\sin ^{2} \,\theta\right)}{\cos \,\theta\left(1-4 \,\sin ^{2} \theta\right)}\right]$
$=\frac{1}{2}[\tan \,27 \,\theta-\tan \,\theta]$
So, $k=\frac{1}{2}$