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If a (tanθ + cot θ ) = 1, sinθ + cos θ = b with 0° < θ < 90°, then a relation between a and b is
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- b² = 2 (a + 1)
- b² = 2 (a – 1)
- 2a = b² – 1
- 2a = b² + 1
Correct Option: C
a (tanθ + cotθ) = 1
| ⇒ a |  | sin θ | + | cos θ |  | = 1 | |
| cos θ | sin θ |
| ⇒ a | | sin² θ + cos² θ | | = 1 | ||
| sin θ.cos θ |
⇒ sinθ . cosθ = a ....(i)
sinθ + cosθ = b
On squaring both sides,
sin²θ + cos²θ + 2 sinθ . cosθ = b²
⇒ 1 + 2a = b²
⇒ 2a = b² –1
