Correct Answer – Option 2 : – 3

Given:

\(\frac{{{{\sin }^2}{}52^\circ + 2 + {{\sin }^2}{{ }}38^\circ }}{{4{}{{\cos }^2}43^\circ – 5 + 4{{\cos }^2}{{}}47^\circ }}\)

Identity used:

Sin θ = cos(90 – θ)

Sin² θ + cos² θ = 1

Calculation:

\(\frac{{{{\sin }^2}{}52^\circ + 2 + {{\sin }^2}{{ }}38^\circ }}{{4{}{{\cos }^2}43^\circ – 5 + 4{{\cos }^2}{{}}47^\circ }}\)

using Sin θ = cos(90 – θ)

⇒ \(\frac{{si{n^2}52^\circ \; + \;2\; + \;si{n^2}\left( {90 – 52} \right)^\circ }}{{4co{s^2}43^\circ – 5\; + \;co{s^2}\left( {90\; – 43} \right)^\circ }}\)

⇒ \(\frac{{si{n^2}52^\circ \; + \;2\; + \;co{s^2}52^\circ }}{{4co{s^2}43^\circ – 5\; + \;si{n^2}43^\circ }}\)

Applying Sin2 θ + cos2 θ = 1

⇒ (1 + 2)/(4 – 5) ⇒ -3

∴ The value of \(\frac{{{{\sin }^2}{}52^\circ + 2 + {{\sin }^2}{{ }}38^\circ }}{{4{}{{\cos }^2}43^\circ – 5 + 4{{\cos }^2}{{}}47^\circ }}\) is -3.