Question

$\text{If}A={340}^{\text{o}},\text{then}$ $\sqrt{1-\sin A}-\sqrt{1+\sin A}$ is equal to

• A. $2\sin {10}^{\text{o}}$
• B. $-2\sin {10}^{\text{o}}$
• C. $2\cos {10}^{\text{o}}$
• D. $-2\cos {10}^{\text{o}}$

Answer 1

The correct option is A $2\sin {10}^{\text{o}}$

As,
$\sin {170}^{\text{o}}+\cos {170}^{\text{o}}>0\because \sin {10}^{\text{o}}+\cos {10}^{\text{o}}>0\sin {170}^{\text{o}}-\cos {170}^{\text{o}}<0\because \sin {10}^{\text{o}}-\cos {10}^{\text{o}}<0$

$A={340}^{\text{o}}\sqrt{1-\sin A}-\sqrt{1+\sin A}=\sqrt{{(\sin \frac{A}{2}-\cos \frac{A}{2})}^2}-\sqrt{{(\sin \frac{A}{2}+\cos \frac{A}{2})}^2}=|\sin \frac{A}{2}-\cos \frac{A}{2}|-|\sin \frac{A}{2}+\cos \frac{A}{2}|=|\sin {170}^{\text{o}}-\cos {170}^{\text{o}}|-|\sin {170}^{\text{o}}+\cos {170}^{\text{o}}|=\sin {170}^{\text{o}}-\cos {170}^{\text{o}}+\sin {170}^{\text{o}}+\cos {170}^{\text{o}}=2\sin {170}^{\text{o}}=2\sin {10}^{\text{o}}$