The value of $(\frac{1-cotθ}{1-tanθ})^2 +1$, if 0° < θ < 90°, is equal to:
Answer & explanation
Correct answer: option 1
( \(\frac{1 - cotθ }{1 + tanθ}\) )² + 1
= ( \(\frac{1 - cosθ/sinθ }{1 + sinθ/cosθ}\) )² + 1
= ( \(\frac{(sinθ - cosθ)² }{sin²θ}\) × ( \(\frac{cos²θ }{ ( cosθ + sinθ )²}\) + 1
= ( \(\frac{cos² θ}{sin²θ}\) + 1
= cot²θ + 1
{ Using , cosec²θ - cot²θ = 1 }
= cosec²θ