If \(\frac{sinx}{siny}\)=\(\frac{1}{2}\) and \(\frac{cosx}{cosy}\)=\(\frac{3}{2}\), then

find the value of cot²x + sin²x.

\(\frac{889}{160}\)

⇒ \(\frac{sinx}{siny}\)=\(\frac{1}{2}\) ⇒ 2sinx = siny  ....(1)

⇒ \(\frac{cosx}{cosy}\)=\(\frac{3}{2}\) ⇒ 2cosx = 3cosy  ...(2)

Squaring and Adding eq. (1) & (2)

4sin²x = sin²y

4cos²x = 9cos²y

⇒ 4sin²x + 4cos²x = sin²y + 9cos²y

4(sin²x + cos²x) = sin²y + 9cos²y

4 = sin²y + cos²y + 8cos²y                            [sin²θ + cos²θ = 1]

4 = 1 + 8cos²y

3 = 8cos²y

\(\frac{3}{8}\) = cos²y

cosy = \(\frac{\sqrt {3}}{\sqrt {8}}\)

Now, if we divide both given equation we get:

tanx . coty= \(\frac{1}{3}\)

tanx × \(\frac{\sqrt {3}}{\sqrt {5}}\) = \(\frac{1}{3}\)

tanx = \(\frac{\sqrt {5}}{3\sqrt {3}}\) = \(\frac{P}{B}\)

H = \(\sqrt {(3\sqrt {3})^2 + (\sqrt {5})^2}\)

H = 4\(\sqrt {2}\)

Put in & find out:

⇒ cot²x + sin²x = (\(\frac{3\sqrt {3}}{\sqrt {5}}\))² + (\(\frac{\sqrt {5}}{4\sqrt {2}}\))² = \(\frac{889}{160}\)