Prove that (1+ cos a )/(1-cos a ) = ((cosec a + cot a ))^2

1.	Prove that (1+ cos a )/(1-cos a ) = ((cosec a + cot a ))^2
Answer» To PROVE: $\Longrightarrow \sf \dfrac{1 + cosA}{1 - cosA} = \big\{cosecA + cotA\big\}^2$ SOLUTION: Taking the LHS we get: $\Longrightarrow \sf \dfrac{1 + cosA}{1 - cosA}$ Let's multiply both the numerator and DENOMINATOR by the conjugate of 1 - cosA which is 1 + cosA. $\Longrightarrow \sf \dfrac{1 + cosA}{1 - cosA} \ \times \ \dfrac{1 + cosA}{1 + cosA}$ $\Longrightarrow \sf \dfrac{\Big\{1 + cosA\Big\}\Big\{1 + cosA\Big\}}{\Big\{1 - cosA\Big\} \Big\{1 + cosA\Big\}}$ Using the below algebraic identities we get: (a + B) (a + b) = (a + b)² (a + b) (a - b) = a² - b² $\Longrightarrow \sf \dfrac{\Big\{1 + cosA\Big\}^2}{\Big\{1\Big\}^2 - \Big\{cosA\Big\}^2}$ Using 1 - cos²A = sin²A we get; $\Longrightarrow \sf \dfrac{\Big\{1 + cosA\Big\}^2}{\Big\{sinA\Big\}^2}$ $\Longrightarrow \sf \Bigg\{\dfrac{1 + cosA}{sinA}\Bigg\}^2$ $\Longrightarrow \sf \Bigg\{\dfrac{1}{sinA} + \dfrac{cosA}{sinA}\Bigg\}^2$ Using the below trigonometric ratios we get; 1/sinA = cosecA cosA/sinA = cotA $\Longrightarrow \sf \big\{ cosecA + cotA \big\}^2$ LHS = RHS Hence PROVED.

Answer»

To PROVE:

$\Longrightarrow \sf \dfrac{1 + cosA}{1 - cosA} = \big\{cosecA + cotA\big\}^2$