Question

The value of $\frac{{{{\sin }^2}{}52^\circ + 2 + {{\sin }^2}{{ }}38^\circ }}{{4{}{{\cos }^2}43^\circ - 5 + 4{{\cos }^2}{{}}47^\circ }}$

• A. 1/3
• B. -3
• C. 3
• D. -1/3

Answer 1

Answer:

B. -3

Solution:

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)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.