.prove that cos (90-A).sinA + sin(90 - A).cosA = 1
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The limit of a(x)/b(x) as x approaches 0 is gotten as;
lim a(x)/b(x); x→0 = 0
- The image showing the semi circle and isosceles triangle is missing and so i have attached it.
- Formula for <em>area</em> of a semi circle is;
a(x) = A = ¹/₂πr²
- <em>Area</em> of triangle is;
b(x) = A = ¹/₂ × base × height
- From the image, ∠PQR = θ
Thus; a(x) = ¹/₂π(10 sin (θ/2))²
a(x) = ¹/₂π(100 sin² (θ/2))
Similarly;
b(x) = ¹/₂(20 sin (θ/2)) × (10 cos (θ/2))
b(x) = 100 sin (θ/2) cos (θ/2)
- Thus;
a(x)/b(x) = ¹/₂π(100 sin² (θ/2)) ÷ 100 sin (θ/2) cos (θ/2)
100 will cancel out and the sin (θ/2) at the denominator will be eradicated
upon further simplification to give;
a(x)/b(x) = ¹/₂π(sin (θ/2)) ÷ cos (θ/2)
In trigonometry, we know that; tan θ = sin θ/cos θ
Thus;
a(x)/b(x) = ¹/₂π(sin (θ/2)) ÷ cos (θ/2) = ¹/₂π(tan (θ/2))
We want to find the limit as θ approaches <em>zero.</em>
Thus;
lim θ → 0 = ¹/₂π(tan (0/2))
⇒ ¹/₂π tan 0
⇒ 0
Read more at; https://brainly.in/question/9387458
