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Mathematics

If sec θ + tan θ = x, then sec θ is equal to :

  1. x2+1x\dfrac{x^2 + 1}{x}

  2. x2−1x\dfrac{x^2 − 1}{x}

  3. x2+12x\dfrac{x^2 + 1}{2x}

  4. x2−12x\dfrac{x^2 − 1}{2x}

Trigonometric Identities

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Answer

⇒ sec θ + tan θ = x ….(1)

We know that,

⇒ sec2 θ − tan2 θ = 1

⇒ (sec θ + tan θ)(sec θ − tan θ) = 1

⇒ x (sec θ − tan θ) = 1

⇒ sec θ − tan θ = 1x\dfrac{1}{x} ….(2)

Adding eqn (1) and (2):

⇒ sec θ + tan θ + sec θ − tan θ = x + 1x\dfrac{1}{x}

⇒ 2 sec θ = x+1xx + \dfrac{1}{x}

⇒ sec θ = 12(x+1x)\dfrac{1}{2}\Big(x + \dfrac{1}{x}\Big)

⇒ sec θ = 12(x2+1x)\dfrac{1}{2}\Big(\dfrac{x^2 + 1}{x}\Big)

⇒ sec θ = x2+12x\dfrac{x^2 + 1}{2x}

Hence, option 3 is the correct option.

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