If x = cotθ + cosy , z = cotθ - cosy, find \(\frac{x^2-z^2}{\sqrt {xz}}\)
Answer & explanation
Correct answer: option 2
Put θ = 45°
y = 45°
x = 1 + \( \frac {1}{\sqrt {2}}\) = \(\frac {\sqrt {2}+1}{\sqrt {2}} \)
z = 1 - \( \frac {1}{\sqrt {2}}\) = \(\frac {\sqrt {2}-1}{\sqrt {2}} \)
Put in find
\(\frac{x^2-z^2}{\sqrt {xz}}\) ⇒ \(\frac{(\frac{\sqrt {2}+1}{\sqrt {2}})^{2} +(\frac{\sqrt {2}-1}{\sqrt {2}})^{2}}{\sqrt {( \frac{\sqrt {2}+1}{\sqrt {2}} ) ( \frac{\sqrt {2}-1}{\sqrt {2}} )}}\) = \(\frac{\frac{2 \sqrt {2}}{\sqrt {2}} \frac{3}{\sqrt {2}}}{\sqrt {1- \frac{1}{2} }}\)
⇒ \(\frac{2\sqrt {2}}{\frac{1}{\sqrt {2}}}\)
= 4