Question

If x = a(secA + tanA)² and y = b(secA - tanA)⁴, then x⁴y² = ​

Answer 1

Answer:

• x = a(secA + tanA)²
• y = b(secA - tanA)⁴
• x⁴y² = ?

First we will find the value of x², as after that only we can find the value of x⁴ :

⇒ x = a(secA + tanA)²

• Squaring both sides

⇒ (x)² = [a(secA + tanA)²]²

⇒ x² = a²(secA + tanA)⁴

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• According to the Question :

• Multiplying both term i.e. x² and y

⇢ x².y = a²(secA + tanA)⁴ × b(secA - tanA)⁴

⇢ x²y = a²b(secA + tanA)⁴(secA - tanA)⁴

• Taking power 4 in common

⇢ x²y = a²b[(secA + tanA)(secA - tanA)]⁴

• (a + b)(a - b) = a² - b²

⇢ x²y = a²b[sec²A - tan²A]⁴

• 1 + tan²A = sec²A ; sec²A - tan²A = 1

⇢ x²y = a²b[1]⁴

⇢ x²y = a²b

• Squaring both sides

⇢ (x²y)² = (a²b)²

⇢ x⁴y² = a⁴b²

∴ Hence, required value of x⁴y² is a⁴b².

Answer 2

Answer:

• The value of x⁴y² is a⁴b².

Step-by-step explanation:

Given that:

• x = a(secA + tanA)²
• y = b(secA - tanA)⁴

To Find:

• The value of x⁴y².

Identities used:

• sec²A - tan²A = 1

Finding the value of x²:

⇒ x = a(secA + tanA)²

• Squaring both sides,

⇒ x² = {a(secA + tanA)²}²

⇒ x² = a²(secA + tanA)⁴

Finding the value of x⁴y²:

⇒ x²y = a²(secA + tanA)⁴•b(secA - tanA)⁴

⇒ x²y = a²b(secA + tanA)⁴•(secA - tanA)⁴

• Taking power common,

⇒ x²y = a²b{(secA + tanA)(secA - tanA)}⁴

• [(a + b) (a - b) = a² - b²]

⇒ x²y = a²b{(sec²A - tan²A}⁴

• [sec²A - tan²A = 1]

⇒ x²y = a²b(1)⁴

⇒ x²y = a²b

• Squaring both sides,

⇒ (x²y)² = (a²b)²

⇒ x⁴y² = a⁴b²