Question

A + B = 90°,
then,
sin² A + sin² B / cos² A+ cos² B​

Answer 1

Answer:

sin²A + sin² B / cos² A + cos² B

=> sin² ( A + B) / cos² ( A + B)

=> sin²90° / cos²90°

=> 1/0

=> 1

Answer 2

Given:

• A + B = 90°

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To find:

• (sin² A + sin² B )/(cos² A+ cos² B)

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Step-by-step explanation:

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We have,

• (sin² A + sin² B )/(cos² A+ cos² B)
• (sin²(A + B))/(cos²(A + B)

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As we are provided that A + B = 90°, so we will substitute its value in the above equation and then using trigonometric table, we will find the value of sin 90° and cos 90°. At last doing basic mathematical calculations, we can easily find out its value.

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• (sin² 90°)/(cos² 90°)

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• sin 90° = 1
• cos 90° = 0

Substituting the values, we have:

• = 1²/0
• = 1/0
• = undefined ∞

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Trigonometric table:

$\Large{ \begin{tabular}{|c|c|c|c|c|c|} \cline{1-6} \theta & \sf 0^{\circ} & \sf 30^{\circ} & \sf 45^{\circ} & \sf 60^{\circ} & \sf 90^{\circ} \\ \cline{1-6} \sin & 0 & \dfrac{1}{2 } & \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} & 1 \\ \cline{1-6} \cos & 1 & \dfrac{ \sqrt{ 3 }}{2} } & \dfrac{1}{ \sqrt{2} } & \dfrac{ 1 }{ 2 } & 0 \\ \cline{1-6} \tan & 0 & \dfrac{1}{ \sqrt{3} } & 1 & \sqrt{3} & \infty \\ \cline{1-6} \cot & \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } &0 \\ \cline{1 - 6} \sec & 1 & \dfrac{2}{ \sqrt{3}} & \sqrt{2} & 2 & \infty \\ \cline{1-6} \csc & \infty & 2 & \sqrt{2 } & \dfrac{ 2 }{ \sqrt{ 3 } } & 1 \\ \cline{1 - 6}\end{tabular}}$

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