Question

prove that cos theta over 1 - tan theta +sin theta over 1 - cot theta is equal to sin theta + cos theta​

Answer 1

$\mathsf{Given :\;\dfrac{cos\theta}{1 - tan\theta} + \dfrac{sin\theta}{1 - cot\theta}}$

$\bigstar\;\;\textsf{We know that : \boxed{\mathtt{cot\theta = \dfrac{cos\theta}{sin\theta}}}}$

$\bigstar\;\;\textsf{We know that : \boxed{\mathtt{tan\theta = \dfrac{sin\theta}{cos\theta}}}}$

$\mathsf{\implies \dfrac{cos\theta}{1 - \dfrac{sin\theta}{cos\theta}} + \dfrac{sin\theta}{1 - \dfrac{cos\theta}{sin\theta}}}$

$\mathsf{\implies \dfrac{cos\theta}{\dfrac{cos\theta - sin\theta}{cos\theta}} + \dfrac{sin\theta}{\dfrac{sin\theta - cos\theta}{sin\theta}}}$

$\mathsf{\implies \dfrac{cos^2\theta}{cos\theta - sin\theta} + \dfrac{sin^2\theta}{sin\theta - cos\theta}}$

$\mathsf{\implies \dfrac{cos^2\theta}{cos\theta - sin\theta} + \dfrac{sin^2\theta}{-(cos\theta - sin\theta)}}$

$\mathsf{\implies \dfrac{cos^2\theta}{cos\theta - sin\theta} - \dfrac{sin^2\theta}{cos\theta - sin\theta}}$

$\mathsf{\implies \dfrac{cos^2\theta - sin^2\theta}{cos\theta - sin\theta}}$

★  We know that : (A + B)(A - B) = A² - B²

$\mathsf{\implies \dfrac{(cos\theta - sin\theta)(cos\theta + sin\theta)}{cos\theta - sin\theta}}$

$\implies \large\boxed{\mathsf{cos\theta + sin\theta}}$