Question

please give solution as fast as possible with solution without giving nonsense ideas ​

Answer 1

Question:

$\sf Evaluate: \dfrac{sin^2 \ 63^\circ + sin^2 \ 27^\circ}{cos^2 \ 17^\circ + cos^2 \ 73^\circ}$

Solution:

We know that,

90° - 63° = 27°

90° - 27° = 63°

90° - 17° = 73°

90° - 73° = 17°

Also,

sinθ = cos(90° - θ)

cosθ = sin(90° - θ)

$\Longrightarrow \ \sf \dfrac{sin^2 \ 63^\circ + sin^2 \ 27^\circ}{cos^2 \ 17^\circ + cos^2 \ 73^\circ}$

$\Longrightarrow \ \sf \dfrac{sin^2(90^\circ - 27^\circ) + sin^2 \ 27^\circ}{cos^2(90^\circ - 73^\circ) + cos^2 \ 73^\circ}$

Using sin(90° - θ) = cosθ and cos(90° - θ) = sinθ we get,

$\Longrightarrow \ \sf \dfrac{cos^2 \ 27^\circ + sin^2 \ 27^\circ}{sin^2 \ 73^\circ + cos^2 \ 73^\circ}$

We know that sin²θ + cos²θ = 1

Applying it above we get,

$\Longrightarrow \ \sf \dfrac{1}{1}$

$\Longrightarrow \ \sf 1$

Therefore,

$\boxed{\sf \dfrac{sin^2 \ 63^\circ + sin^2 \ 27^\circ}{cos^2 \ 17^\circ + cos^2 \ 73^\circ} = 1}$

Answer 2

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