Question

$\left(\frac{\sin 3\theta -\cos 3\theta }{\sin \theta +\cos \theta }\right)+1=$

• A. $2\sin 2\theta$
• B. $2\cos 2\theta$
• C. $\tan 2\theta$
• D. $\cot 2\theta$

Answer 1

The correct option is A

$2\sin 2\theta$

Explanation for the correct answer:

Simplifying the equations using trigonometric identities:

$$\begin{gathered} =\left(\frac{\sin 3\theta -\cos 3\theta }{\sin \theta +\cos \theta }\right)+1 \\ =\left(\frac{3\sin \theta –4{\sin }^3\theta –4{\cos }^3\theta +3\cos \theta }{\sin \theta +\cos \theta }\right)+1\left[\because \sin 3\theta =3\sin \theta -4{\sin }^3\theta and\cos 3\theta =4{\cos }^3\theta +3\cos \theta \right] \\ =\left(\frac{3\left(\sin \theta +\cos \theta \right)-4\left({\sin }^3\theta +{\cos }^3\theta \right)}{\sin \theta +\cos \theta }\right)+1[\mathrm{Getting}\mathrm{the}\mathrm{common}\mathrm{terms}\mathrm{together}] \end{gathered}$$

$\left[\because a^3+b^3=(a+b)(a^2+b^2–ab)\right]$, applying this on the above

$$\begin{gathered} =\left(\frac{3\left(\sin \theta +\cos \theta \right)-4\left(\left(\sin \theta +\cos \theta \right)\left({\sin }^2\theta +{\cos }^2\theta +\sin \theta \cos \theta \right)\right)}{\sin \theta +\cos \theta }\right)+1 \\ =\left(\frac{3\left(\sin \theta +\cos \theta \right)-4\left(\left(\sin \theta +\cos \theta \right)\left(1+\sin \theta \cos \theta \right)\right)}{\sin \theta +\cos \theta }\right)+1 \\ =\left[\frac{\left(\sin \theta +\cos \theta \right)\left(3–4+4\sin \theta \cos \theta \right)}{\sin \theta +\cos \theta }\right]+1 \\ =-1+4\sin \theta \cos \theta +1 \\ =2(2\sin \theta \cos \theta ) \\ =2\sin 2\theta \end{gathered}$$

Therefore, option (A) is correct.