Question

$(2n + 1)\frac{\pi}{2}$then (rc) $n = - 2, - 1,0,1,2$

• A. $\therefore$
• B. $x = \frac{- 3\pi}{4},\frac{- \pi}{4},\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4}$
• C. $\frac{- 3\pi}{2},\frac{- \pi}{2},\frac{\pi}{2}, ⥂ \frac{3\pi}{2},\frac{5\pi}{2}$
• D. None of these

Answer 1

Answer: (B)

$x = \frac{- 3\pi}{4},\frac{- \pi}{4},\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4}$

Answer 2

We have $(2n \pm 1)\frac{\pi}{4}$

or $\frac{2\tan x + 1 - \tan^{2}x}{1 + \tan^{2}x} = \tan x$

$\Rightarrow$ $\tan^{3}x + \tan^{2}x - \tan x - 1 = 0$,

$\Rightarrow (\tan^{2}x - 1)(\tan x + 1) = 0$

$\Rightarrow x = m\pi \pm \frac{\pi}{4}$

Now put $n = 0$

$\theta = - \frac{\pi}{2}\theta = \frac{\pi}{4}$and $r$

Since $\therefore$, therefore $\sin\theta = \frac{1}{2}, - \frac{3}{2}$ only.