1.

If (a² − b²) sinθ + 2ab cosθ = a² + b², then the value of secθ is:1). 1/2 × (a² + b²)2). 1/2ab × (a² – b²)3). 1/2ab × (a² + b²)4). 1/ab × (a² + b²)

Answer»

Dividing the whole equation by cosθ,

⇒ (a² − b²)tanθ + 2AB = (a² + b²)secθ (? 1/cosθ = secθ)

⇒ 2ab = (a² + b²)secθ – (a² − b²)tanθ

⇒ 2ab = a² (secθ – tanθ) + b² (secθ + tanθ)

Dividing the whole by ab,

⇒ 2 = (a/b)(secθ – tanθ) + (b/a)(secθ + tanθ) ………… (1)

Now,

⇒ sec²θ – tan²θ = 1

⇒ (secθ + tanθ) × (secθ – tanθ) = 1

⇒ (secθ + tanθ) = 1/(secθ – tanθ)

Substitute above in (1),

⇒ 2 = (a/b)(secθ – tanθ) + (b/a) × 1/(secθ – tanθ)

Let (a/b)(secθ – tanθ) = Z

⇒ 2 = z + 1/z

⇒ 2z = z² + 1

⇒ z² – 2z + 1 = 0

⇒ (z – 1)² = 0

⇒ z = 1

⇒ (a/b) (secθ – tanθ) = 1

⇒ (secθ – tanθ) = b/a

⇒ (secθ + tanθ) = a/b (? it is INVERSE of (sec θ – TAN θ))

Solving above,

⇒ 2secθ = b/a + a/b

⇒ 2secθ = (a² + b²)/ab

∴ secθ = 1/2 ab × (a² + b²)


Discussion

No Comment Found