Question

(cosec x - cotx)^2 = 1 - cos x / 1 + cos x​

Answer 1

To Prove:

$\Longrightarrow \rm \bigg[cosecx - cotx\bigg]^{2} = \dfrac{1 - cosx}{1 + cosx}$

Take the LHS:

$\Longrightarrow \rm \bigg[cosecx - cotx \bigg]^{2}$

Using (a - b)² = a² + b² - 2ab we get:

$\Longrightarrow \rm cosec^{2}x + cot^{2}x - 2\big[cosecx\big]\big[cotx\big]$

In the RHS, all the terms are in terms of cosx, so let's try to express the equation we've got above in terms of cosx.

Substitute the following:

⇒ 1/cosec²x = sin²x

⇒ 1/cot²x = cos²x/sin²x

⇒ cosecx = 1/sinx

⇒ cotx = cosx/sinx

$\Longrightarrow \rm \dfrac{1}{sin^2x} + \dfrac{cos^2x}{sin^2x} - 2\bigg[\dfrac{1}{sinx}\bigg]\bigg[\dfrac{cosx}{sinx} \bigg]$

$\Longrightarrow \rm \dfrac{1}{sin^2x} + \dfrac{cos^2x }{sin^2x} - \dfrac{2cosx}{sin^2x}$

$\Longrightarrow \rm \dfrac{1 + cos^2x - 2cosx}{sin^2x}$

Here, let's apply  (a - b)² = a² + b² - 2ab.

[a = 1 ; b = cosx]

$\Longrightarrow \rm \dfrac{\ \big[1 - cosx\big]^2}{sin^2x}$

Apply sin²x = 1 - cos²x.

$\Longrightarrow \rm \dfrac{\ \big[1 - cosx\big]^2}{1 - cos^2x}$

Apply a² - b² = (a + b)(a - b).

$\Longrightarrow \rm \dfrac{\ \big[1 - cosx\big]^2}{\big[1 - cosx\big] \big[1 + cosx\big]}$

Divide [1 - cosx] and [1 - cosx]².

$\Longrightarrow \rm \dfrac{1 - cosx}{1 + cosx}$

LHS = RHS

Hence proved.