Question

ans this trigonometry identities que. geniuses​

Answer 1

Given

a³ = cosec∅ - sin∅

b³ = sec∅ - cos∅

⇒ a³ = 1/sin∅ - sin∅ (since, cosec∅ = 1/sin∅)

⇒ a³ = (1 - sin²∅)/sin∅

⇒ a³ = cos²∅/sin∅  (Since, cos²∅ + sin²∅ = 1)

⇒ a = $\sf\sqrt[3]{\frac{cos^2\theta}{sin\theta}}$

Similarly,

b³ = 1/cos∅ - cos∅

⇒ b³ = (1 - cos²∅)/cos∅

⇒ b³ = sin²∅/cos∅

⇒ b =  $\sf\sqrt[3]{\frac{sin^2\theta}{cos\theta}}$

Now,

a²b²(a² + b²) can be written as

⇒ ab(a³b + ab³)

Substitue values of a and b

$\sf ab = \sqrt[3]{\frac{cos^2\theta}{sin\theta}\times\frac{sin^2\theta}{cos\theta}} \\\\\implies ab = \sqrt[3]{cos\theta sin\theta}\\\\a^3b = \frac{cos^2\theta}{sin\theta}\times ({\frac{sin^2\theta}{cos\theta}})^{\frac{1}{3}}\\\\\implies a^3b = \frac{cos^{2-\frac{1}{3}}\theta}{sin^{1-\frac{2}{3}}\theta}\\\\\implies a^3b = \frac{cos^{\frac{5}{3}}\theta}{sin^{\frac{1}{3}}\theta}$

Similarly,

$\sf ab^3 = \frac{sin^{\frac{5}{3}}\theta}{cos^{\frac{1}{3}}\theta}\\\\so \\\\a^2b^2(a^2 + b^2)\\\\= ab(a^3b + ab^3)$

$\sf(\sqrt[3]{cos\theta sin\theta})(\frac{cos^{\frac{5}{3}}\theta}{sin^{\frac{1}{3}}\theta}+ \frac{sin^{\frac{5}{3}}\theta}{cos^{\frac{1}{3}}\theta})\\\\\implies (\sqrt[3]{cos\theta sin\theta})(\frac{cos^2\theta + sin^2\theta}{\sqrt[3]{cos\theta sin\theta}})\\\\(on\ taking\ LCM) \\\\\implies cos^2\theta + sin^2\theta\\\\(after\ cancelling\ \sqrt[3]{cos^2\theta sin^2\theta})\\\\\implies 1 \\\\(since, cos^2\theta + sin^2\theta = 1)$

Hence proved