1.

If xcosθ – sinθ = 1 then x2 + (1 + x2)sinθ equals :1). 02). 3sec2θ (1 + sinθ) + 43). 4sec2θ (1 + sinθ)4). 4sec2θ (1 + sinθ) - 3

Answer»

We know that,

sec2θ - tan2θ = 1

Given, xcos θ – sinθ = 1

⇒ xcosθ = 1 + sinθ

Dividing the above EXPRESSION by cosθ

X = secθ + tanθ

Squaring both SIDES, we get

⇒ x2 = (secθ + tanθ)2

⇒ x2 = sec2θ + tan2θ + 2secθ tanθ

⇒ x2 = 1 + tan2θ + tan2θ + 2secθ tanθ

⇒ x2 = 1 + 2 tan2θ + 2secθ tanθ . . . . . . . eq (1)

Adding 1 on both sides, we get

⇒ 1 + x2 = 2 + 2tan2θ + 2secθtanθ

⇒ (1 + x2) sinθ = 2sinθ + 2tan2θsinθ + 2secθtanθsinθ

⇒ (1 + x2) sinθ = 2sinθ + 2tan2θsinθ + 2tan2θ

⇒ x2 + (1 + x2)sinθ = 1 + 2 tan2θ + 2secθ tanθ + 2sinθ + 2tan2θ sinθ + 2tan2θ (from eq (1))

⇒ x2 + (1 + x2)sinθ = 4 tan2θ + 1 + 2sinθ + 2tan2θ sinθ + 2secθ tanθ

Using sec2θ – 1 = tan2θ, tanθ = sinθ/cosθ and secθ = 1/cosθ

 ⇒ x2 + (1 + x2)sinθ = 4sec2θ -4 + 1 + 2sinθ + 2 sec2θsinθ – 2sinθ + 2sec2θsinθ

⇒ x2 + (1 + x2)sinθ = 4sec2θ (1 + sinθ) - 3



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