If secθ + tanθ = x, then find sinθ.

Correct — A. Start from sec²θ − tan²θ = 1 and factorise it.

(secθ + tanθ)(secθ − tanθ) = 1
secθ + tanθ = x is given, so secθ − tanθ = 1⁄x

Add the two: 2secθ = x + 1⁄x, so secθ = (x² + 1)⁄2x
Subtract them: 2tanθ = x − 1⁄x, so tanθ = (x² − 1)⁄2x

sinθ = tanθ ⁄ secθ, and the 2x cancels:
sinθ = (x² − 1)⁄(x² + 1)

That is the fraction on option (a), printed as (x² − 1)⁄(1 + x²).

Why the others are wrong

• (b)
  Option (b) shows (x² + 1)⁄(1 − x²) — the right fraction inverted and sign-flipped, which is −cosec θ. Its size exceeds 1 whenever x is greater than 1, and sin θ never exceeds 1 in magnitude, so it can be rejected on size alone.
• (c)
  Option (c) shows (x² − 1)⁄(1 + 2x²). The numerator is right, so the slip is in sec θ: x + 1⁄x gives (x² + 1)⁄2x, and that 2 belongs to the denominator that cancels, not to the x² of the final fraction.
• (d)
  Option (d) shows (1 − x²)⁄(1 + x²) — the key with its numerator's sign reversed. It comes from writing 2 tan θ = 1⁄x − x instead of x − 1⁄x when the two equations are subtracted.

One identity does all the work: sec²θ − tan²θ = 1, which factorises as (secθ + tanθ)(secθ − tanθ) = 1.

So the moment secθ + tanθ = x is given, its partner is forced: secθ − tanθ = 1⁄x. That leaves two linear equations in secθ and tanθ, which separate by adding and subtracting.

Finish with sinθ = tanθ ⁄ secθ — tan is sin⁄cos and sec is 1⁄cos, so the cos cancels. The 2x denominators cancel too, leaving a clean ratio in x.

Key facts

• sec²θ − tan²θ = 1, so if sec θ + tan θ = x then sec θ − tan θ = 1⁄x.
• sin θ = tan θ ⁄ sec θ, since tan θ = sin θ ⁄ cos θ and sec θ = 1 ⁄ cos θ.
• For sec θ + tan θ = x: sec θ = (x² + 1)⁄2x and tan θ = (x² − 1)⁄2x.

Common traps

• Squaring sec θ + tan θ = x and wrestling with the cross term instead of using the conjugate.
• Losing the sign when subtracting the two equations, which lands on option (d)'s fraction.
• Answering with sec θ or cosec θ after correctly finding both intermediate values.

No directly related past PYQ was found.