If $sin\theta + cos\theta$ = $\sqrt{2} sin (9O^{0} - \theta)$ then $cot\theta $ is equal to :
1). $\sqrt{2}$
2). 0
3). $\sqrt{2}-1$
4). $\sqrt{2}+1$
sin$\theta$ + cos$\theta$ = $\sqrt{2}$ sin (90° – $\theta$)
=> sin$\theta$ + cos$\theta$ = $\sqrt{2}$ cos$\theta$
=>$\sqrt{2}$ cos$\theta$ – cos$\theta$ = sin$\theta$
=> cos$\theta$ ( $\sqrt{2}-1$)= sin$\theta$
=> $\frac{sin\theta}{cos\theta}$ = $\frac{1}{\sqrt{2}-1}$
=> $cot\theta$ = $\frac{\sqrt{2}+1}{2-1}$ = $\sqrt{2}+1$ .