Question

If ${\tan }^{2}2\theta =co{t}^{2}a$ then the general equation is

• A. $\theta =\left(\frac{1}{4}\right)\left\{n\pi \pm \left(\frac{\pi }{2}-a\right)\right\},\forall n\in z$
• B. $\theta =\left(\frac{1}{2}\right)\left\{n\pi \pm \left(\frac{\pi }{4}-a\right)\right\},\forall n\in z$
• C. $\theta =\left(\frac{1}{2}\right)\left\{n\pi \pm \left(\frac{\pi }{2}-a\right)\right\},\forall n\in z$
• D. $\theta =\left(\frac{1}{4}\right)\left\{n\pi \pm \left(\frac{\pi }{2}+a\right)\right\},\forall n\in z$

Answer 1

The correct option is A $\theta =\left(\frac{1}{4}\right)\left\{n\pi \pm \left(\frac{\pi }{2}-a\right)\right\},\forall n\in z$

$\tan {}^{2}2\theta =\cot {}^{2}a\Rightarrow \tan 2\theta =\cot a\Rightarrow \frac{2\tan \theta }{1-\tan \theta }=\cot a\Rightarrow \frac{\frac{2}{\cot \theta }}{1-\frac{1}{\cot {}^2\theta }}=\cot a\Rightarrow \frac{2\cot \theta }{\cot {}^2\theta -1}=\cot a\Rightarrow \frac{1}{\cot 2\theta }=\cot a\Rightarrow a={\cot }^{-1}\left(\frac{1}{\cot 2\theta }\right)\Rightarrow \theta =\frac{1}{4}\{n\pi +(\frac{\pi }{2}-a\left)\right\},\forall nϵz$