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Evaluate $\tan ^{-1}\bigg(\dfrac{3 \sin 2 \alpha}{5+3 \cos 2 \alpha}\bigg)+\tan ^{-1}\bigg(\dfrac14 \tan \alpha\bigg)$ where $\dfrac{-\pi}{2}<\alpha<\dfrac{\pi}{2}$

I could not approach this problem at all. Please help.

Swadhin
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1 Answers1

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Hint: Using Weierstrass substitution,

$$\dfrac{3\sin2\alpha}{5+3\cos2\alpha}=\dfrac{6\tan\alpha}{5(1+\tan^2\alpha)+3(1-\tan^2\alpha)}=\dfrac{\tan\alpha-\dfrac{\tan\alpha}4}{1+\tan\alpha\cdot\dfrac{\tan\alpha}4}$$

$$\implies\tan^{-1}\dfrac{3\sin2\alpha}{5+3\cos2\alpha}=\tan^{-1}(\tan\alpha)-\tan^{-1}\left(\dfrac{\tan\alpha}4\right)=\alpha-\tan^{-1}\left(\dfrac{\tan\alpha}4\right)$$

lab bhattacharjee
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  • This holds true as $\arctan(-x)=-\arctan x$ and see my answer here : https://math.stackexchange.com/questions/1837410/inverse-trigonometric-function-identity-doubt-tan-1x-tan-1y-pi-tan – lab bhattacharjee Dec 30 '17 at 14:03