InterviewSolution
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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) |
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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) |
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