Question

$\frac{\text{sin}\theta \text{cos}({90}^{\circ }-\theta )\text{cos}\theta }{\text{sin}({90}^{\circ }-\theta )}+\frac{\text{cos}\theta \text{sin}({90}^{\circ }-\theta )\text{sin}\theta }{\text{cos}({90}^{\circ }-\theta )}=$ ___

• A. - 1
• B. 0
• C. 1
• D. 2

Answer 1

The correct option is C

1

$\frac{\text{sin}\theta \text{cos}({90}^{\circ }-\theta )\text{cos}\theta }{\text{sin}({90}^{\circ }-\theta )}+\frac{\text{cos}\theta \text{sin}({90}^{\circ }-\theta )\text{sin}\theta }{\text{cos}({90}^{\circ }-\theta )}$

$=\frac{\text{sin}\theta \left(\text{sin}\theta \right)\text{/}\text{cos}\theta }{\text{/}\text{cos}\theta }+\frac{\text{cos}\theta \left(\text{cos}\theta \right)\text{/}\text{sin}\theta }{\text{/}\text{sin}\theta }$

$={\text{sin}}^2\theta +{\text{cos}}^2\theta$

$=1$