Question

Choose the correct option of general solutions -
$$\begin{array}{cc}\text{Trigonometric Equations}& \text{General Solutions}\\ 1.{\sin }^2\theta ={\sin }^{2}x& A.2n\pi \pm \alpha \\ 2.{\cos }^2\theta ={\cos }^{2}x& B.n\pi +(-1{)}^n\alpha \\ 3.{\tan }^2\theta ={\tan }^{2}x& C.n\pi \pm \alpha \\ & \end{array}$$

• A. $1-C,2-C,3-C$
• B. $1-A,2-A,3-A$
• C. $1-B,2-A,3-C$
• D. $1-A,2-B,3-C$

Answer 1

The correct option is A

$1-C,2-C,3-C$

1 . ${\sin }^2\theta ={\sin }^2\alpha$

$\Rightarrow {\sin }^2\theta -{\sin }^2\alpha =0$

Multiplying $2$ on both sides.

$\Rightarrow 2{\sin }^2\theta =2{\sin }^2\alpha$

So, $1-\cos 2\theta =1-\cos 2\alpha$

$\Rightarrow \cos 2\theta =\cos 2\alpha$

$\theta =n\pi -\alpha$ or $\theta =n\pi +\alpha$

Combining both the equations,

We get, $\theta =n\pi \pm \alpha$

2. ${\cos }^2\theta ={\cos }^2\alpha$

$\Rightarrow {\cos }^2\theta -{\cos }^2\alpha =0$

$\Rightarrow \frac{\cos 2\theta +1}{2}=\frac{\cos 2\alpha +1}{2}$

$\Rightarrow \cos 2\theta =\cos 2\alpha$

$\theta =n\pi -\alpha$ or $\theta =n\pi +\alpha$

Combining both the equations,

We get, $\theta =n\pi \pm \alpha$

3. ${\tan }^2\theta ={\tan }^2\alpha$

$\Rightarrow \frac{{\sin }^2\theta }{{\cos }^2\theta }$ - $\frac{{\sin }^2\alpha }{{\cos }^2\alpha }=0$

$\Rightarrow \frac{{\sin }^2\theta .{\cos }^2\alpha -{\cos }^2\theta .{\sin }^2\alpha }{{\cos }^2\theta .{\cos }^2\alpha }=0$

$\Rightarrow \frac{(\sin \theta .\cos \alpha {)}^2-(\cos \theta .\sin \alpha {)}^2}{{\cos }^2\theta .{\cos }^2\alpha }=0$

$\Rightarrow \frac{(\sin \theta .\cos \alpha +\cos \theta .\sin \alpha )(\sin \theta .\cos \alpha -\cos \theta .\sin \alpha )}{{\cos }^2\theta .{\cos }^2\alpha }=0$

$\Rightarrow \sin (\theta +\alpha )\sin (\theta -\alpha )=0$ where $\theta \text{and}\alpha$ should not be equal to $(2n+1)$$\frac{\pi }{2}n\in Z$

$\Rightarrow \sin (\theta +\alpha )=0\text{or}\sin (\theta -\alpha )=0$

$\theta +\alpha =n\pi$ or $\theta -\alpha =n\pi$

$\theta =n\pi -\alpha$ or $\theta =n\pi +\alpha$

Combining both the equations,

We get, $\theta =n\pi \pm \alpha$

Where $n\in Z,\alpha \text{and}\theta$ should not be equal odd multiples of $\frac{\pi }{2}$.

$\alpha \text{and}\theta \ne (2n+1)\pi /2\in Z$

Hence, the correct answer is Option d.