Question

Prove that
1+cosA/1-cosA - 1-cosA/1+cosA = 4cotA cosecA

Answer 1

Answer:

Step-by-step explanation:

L.H.S:

        (1+CosA)² - (1-CosA)²

=>     -----------------------------

        (1-CosA) (1+CosA)

=>     4CosA

         -------------

          Sin²A

=> 4. CosA/SinA *  1/SinA

=> 4 CotA.CosecA

= R.H.S

Answer 2

To prove: $\dfrac{1+\cos A}{1-\cos A} -\dfrac{1-\cos A}{1+\cos A} = 4 \cot A \text{ cosec A}$

Step-by-step explanation:

Now Consider L.H.S.

We have

$\dfrac{1+\cos A}{1-\cos A} -\dfrac{1-\cos A}{1+\cos A} \\\\= \dfrac{(1+\cos A)^2-(1-\cos A)^2}{(1-\cos A)(1+\cos A)}$

As we know

$(a+b)^2=a^2+b^2+2ab \text { and } (a-b)(a+b)=a^2-b^2$  we have

$\dfrac{1+\cos^2 A+2\cos A- (1+\cos^2 A-2\cos A)}{((1)^2-(\cos A)^2} \\\\=\dfrac{1+\cos^2 A+2\cos A- 1-\cos^2 A+2\cos A}{1-\cos^2 A}$

As $\sin^2\theta +\cos ^2\theta =1$

$=\dfrac{4 \cos A}{\sin^2A} =4 \dfrac{\cos A}{\sin A\times \sin A} =4 \cot A\text { cosec A}$

which is equal to R.H.S .

Hence, proved the required result

i.e. $\dfrac{1+\cos A}{1-\cos A} -\dfrac{1-\cos A}{1+\cos A} = 4 \cot A \text{ cosec A}$