Question

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

Answer 1

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 2

✧ $\sf \dfrac{1 + cosA}{1 - cosA} = (cosecA + cotA)^{2}$

LHS:-

$\sf \dfrac{1 + cosA}{1 - cosA}$

By rationalisation:-

$\sf \dfrac{1 + cosA}{1 - cosA} \times \dfrac{1 + cosA}{1 + cosA}$

$\sf ➝ \: \dfrac{(1 + cosA)(1 + cosA)}{(1 - cosA)(1 + cosA)}$

$\sf \color{fuchsia}By \: using \: algebraic \: identities:-$

$\sf \color{fuchsia} (a + b) (a + b) = (a + b)²$

$\sf \color{fuchsia} (a + b) (a - b) = a² - b²$

$\sf ➝ \: \dfrac{(1 + cosA)^{2}}{{(1)}^{2} - {(cosA)}^{2}}$

$\sf ➝ \: \dfrac{(1 + cosA)^{2}}{1 - cos²A}$

$\sf \color{fuchsia} Now, \: we \: know \: that \: 1 - cos²A = sin²A$

$\sf ➝ \: \dfrac{(1 + cosA)^2}{sin²A}$

$\sf ➝ \: (\dfrac{1 + cosA}{sinA})^{2}$

$\sf ➝ \: (\dfrac{1}{sinA} + \dfrac{cosA}{sinA})^{2}$

$\sf \color{fuchsia} As, \: We \: know \: that,$

$\sf \color{fuchsia} \dfrac{1}{sinA} = cosecA$

$\sf \color{fuchsia} \dfrac{cosA}{sinA} = cotA$

$\sf ➝ \: (\dfrac{1}{sinA} + \dfrac{cosA}{sinA})^{2} = (cosecA + cotA)^{2}$

RHS:-

$\sf (cosecA + cotA)^{2}$

$\sf \therefore LHS = RHS$

$\sf Hence, \: Proved$