Answer :
Answer:
[tex]\dfrac{sin(90^{\circ} - \alpha) \times tan (-\alpha) \times cos(\alpha - 180^{\circ})}{cos(100^{\circ}) \times sin(135 ^{\circ})} = \dfrac{\sqrt{2} \times sin(2 \times \alpha) }{2 \times cos(100^{\circ}) }[/tex]
Explanation:
The given expression is presented as follows;
[tex]\dfrac{sin(90^{\circ} - \alpha) \times tan (-\alpha) \times cos(\alpha - 180^{\circ})}{cos(100^{\circ}) \times sin(135 ^{\circ})}[/tex]
The sine, cosine, and tangent of the common angles are given as follows;
sin(A - B) = sin(A)·cos(B) - cos(A)·sin(B)
∴ sin(90° - α) = sin(90°)·cos(α) - cos(90°)·sin(α) = cos(α)
tan(-α) = -tan(α)
cos(A - B) = cos(A)·cos(B) + sin(A)·sin(B)
∴ cos(α - 180°) = cos(α)·cos(180°) + sin(α)·sin(180°) = -cos(α)
sin(135°) = sin(90° + 45°) = sin(90°)·cos(45°) + sin(45°)·cos(90°) = cos(45°) = √2/2
Therefore, we get;
[tex]\dfrac{sin(90^{\circ} - \alpha) \times tan (-\alpha) \times -cos( 180^{\circ})}{cos(100^{\circ}) \times sin(135 ^{\circ})} = \dfrac{cos( \alpha) \times -tan (\alpha) \times -cos(\alpha )}{cos(100^{\circ}) \times \dfrac{\sqrt{2} }{2}}[/tex]
[tex]\dfrac{cos( \alpha) \times -\dfrac{sin(\alpha)}{cos(\alpha)} \times -cos(\alpha )}{cos(100^{\circ}) \times \dfrac{\sqrt{2} }{2}} = \dfrac{sin( \alpha) \times cos(\alpha )}{cos(100^{\circ}) \times \dfrac{\sqrt{2} }{2}} = \dfrac{2}{\sqrt{2} } \times \dfrac{sin( \alpha) \times cos(\alpha )}{cos(100^{\circ}) }[/tex]
[tex]\dfrac{2}{\sqrt{2} } \times \dfrac{sin( \alpha) \times cos(\alpha )}{cos(100^{\circ}) } = \dfrac{sin(2 \times \alpha) }{\sqrt{2} \times cos(100^{\circ}) }[/tex]
Therefore;
[tex]\dfrac{sin(90^{\circ} - \alpha) \times tan (-\alpha) \times cos(\alpha - 180^{\circ})}{cos(100^{\circ}) \times sin(135 ^{\circ})} = \dfrac{\sqrt{2} \times sin(2 \times \alpha) }{2 \times cos(100^{\circ}) }[/tex]