Question

1+cot alpha + cosec alpha divided by 1-cot alpha +cosec alpha=

Answer 1

$\underline{\textbf{To prove:}}$

$\mathsf{\dfrac{1+cot\,\alpha+cosec\,\alpha}{1-cot\,\alpha+cosec\,\alpha}=\dfrac{1+cos\,\alpha}{sin\,\alpha}}$

$\underline{\textbf{Solution:}}$

$\mathsf{Consider,}$

$\mathsf{\dfrac{1+cot\,\alpha+cosec\,\alpha}{1-cot\,\alpha+cosec\,\alpha}}$

$\textsf{Using the identity,}$

$\boxed{\bf\,cosec^2\theta-cot^2\theta=1}$

$\mathsf{=\dfrac{(cosec^2\alpha-cot^2\alpha)+cot\,\alpha+cosec\,\alpha}{1-cot\,\alpha+cosec\,\alpha}}$

$\textsf{Using the identity,}$

$\boxed{\bf\,a^2-b^2=(a-b)(a+b)}$

$\mathsf{=\dfrac{(cosec\,\alpha-cot\,\alpha)(cosec\,\alpha+cot\,\alpha)+cot\,\alpha+cosec\,\alpha}{1-cot\,\alpha+cosec\,\alpha}}$

$\mathsf{=\dfrac{(cosec\,\alpha+cot\,\alpha)((cosec\,\alpha-cot\,\alpha)+1)}{1-cot\,\alpha+cosec\,\alpha}}$

$\mathsf{=\dfrac{(cosec\,\alpha+cot\,\alpha)(1-cot\,\alpha+cosec\,\alpha)}{1-cot\,\alpha+cosec\,\alpha}}$

$\mathsf{=cosec\,\alpha+cot\,\alpha}$

$\mathsf{=\dfrac{1}{sin\,\alpha}+\dfrac{cos\,\alpha}{sin\,\alpha}}$

$\mathsf{=\dfrac{1+cos\,\alpha}{sin\,\alpha}}$

$\implies\boxed{\bf\dfrac{1+cot\,\alpha+cosec\,\alpha}{1-cot\,\alpha+cosec\,\alpha}=\dfrac{1+cos\,\alpha}{sin\,\alpha}}$