Question

show that (cosec¶-cot¶)^2=1-cos¶/1+cos¶​

Answer 1

Answer:

cosec is 1÷cos so minus and plus cancel

Answer 2

ANSWER:

To Prove:

• (cosec ө - cot ө)² = (1 - cos ө)/(1 + cos ө)

Proof:

We need to prove that,

⇒ (cosec ө - cot ө)² = (1 - cos ө)/(1 + cos ө)

$\implies(\csc\theta-\cot\theta)^2=\dfrac{1-\cos\theta}{1+\cos\theta}$

Taking and solving LHS,

$\implies(\csc\theta-\cot\theta)^2$

We know that,

⇒ cosec A = 1/sin A

And,

⇒ cot A = cos A/sin A

So,

$\implies\left(\dfrac{1}{\sin\theta}-\dfrac{\cos\theta}{\sin\theta}\right)^2$

$\implies\left(\dfrac{1-\cos\theta}{\sin\theta}\right)^2$

So,

$\implies\dfrac{(1-\cos\theta)^2}{(\sin\theta)^2}$

$\implies\dfrac{(1-\cos\theta)^2}{\sin^2\theta}$

We know that,

⇒ sin²A = 1 - cos²A

So,

$\implies\dfrac{(1-\cos\theta)^2}{\sin^2\theta}$

$\implies\dfrac{(1-\cos\theta)^2}{1-\cos^2\theta}$

We know that,

⇒ a² - b² = (a + b)(a - b)

Hence,

$\implies\dfrac{(1-\cos\theta)^2}{1^2-\cos^2\theta}$

$\implies\dfrac{(1-\cos\theta)^2}{(1-\cos\theta)(1+\cos\theta)}$

Cancelling (1 - cos ө) in the fraction,

$\implies\bf\dfrac{1-\cos\theta}{1+\cos\theta}$

As, LHS = RHS,

HENCE PROVED.!!!