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If $sin\theta + cos\theta$ = $\sqrt{2} sin (9O^{0} - \theta)$ then $cot\theta $ is equal to :

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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). $\sqrt{2}+1$

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2 answers

2 vote
Answered by on | Votes 2 |
Correct answer is: $\sqrt{2}+1$

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3 vote
Answered by on | Votes 3 |

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$\theta$

=> $\frac{sin\theta}{cos\theta}$ = $\frac{1}{\sqrt{2}-1}$

=> $cot\theta$ = $\frac{\sqrt{2}+1}{2-1}$ = $\sqrt{2}+1$ .




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