Question

The general values of q for which $\theta =ta{n}^{-1}\left(2ta{n}^2\theta \right)-\frac{1}{2}si{n}^{-1}\left(\frac{3sin2\theta }{5+4cos2\theta }\right)$ holds true,

• A. $\theta =n\pi$
• B. $\theta =n\pi +\frac{\pi }{4}$
• C. $\theta =n\pi +ta{n}^{-1}(-2)$
• D. $\theta =n\pi +ta{n}^{-1}\left(2\right)$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $\theta =n\pi$
B $\theta =n\pi +\frac{\pi }{4}$
C $\theta =n\pi +ta{n}^{-1}(-2)$

$\frac{3sin2\theta }{5+4cos2\theta }=\frac{3.\frac{2tan\theta }{1+ta{n}^2\theta }}{5+4.\frac{1-ta{n}^2\theta }{1+ta{n}^2\theta }}=\frac{6tan\theta }{9+ta{n}^2\theta }=\frac{2\frac{tan\theta }{3}}{1+{\left(\frac{tan\theta }{3}\right)}^2}\theta =ta{n}^{-1}\left(2ta{n}^2\theta \right)-\frac{1}{2}si{n}^{-1}\left(\frac{2\left(\frac{tan\theta }{3}\right)}{1+{\left(\frac{tan\theta }{3}\right)}^2}\right)\theta =ta{n}^{-1}\left(2ta{n}^2\theta \right)-\frac{1}{2}\times 2\times ta{n}^{-1}\left(\frac{tan\theta }{3}\right)\{\because 2ta{n}^{-1}x=si{n}^{-1}\left(\frac{2x}{1+x^2}\right)\theta =ta{n}^{-1}\left(2ta{n}^2\theta \right)-ta{n}^{-1}\left(\frac{tan\theta }{3}\right)\theta =ta{n}^{-1}\left(\frac{2ta{n}^2\theta -\frac{tan\theta }{3}}{1+2ta{n}^2\theta .\frac{tan\theta }{3}}\right)tan\theta =\frac{6ta{n}^2\theta -tan\theta }{3+2ta{n}^3\theta }tan\theta (\frac{6tan\theta -1}{3+2ta{n}^2\theta }-1)=0tan\theta (tan\theta -1)(2ta{n}^2\theta +2tan\theta -4)=0tan\theta =0ortan\theta =1orta{n}^2\theta +tan\theta -2=0\Rightarrow \theta =n\pi or\theta =n\pi +\frac{\pi }{4}or\theta =n\pi +ta{n}^{-1}(-2)$