Question

SinA/CosecA+ cotA=2+SinA/cotA-cosecA​

Answer 1

To prove:-

$\rm \dfrac{sin \ A}{cosec \ A+ cot \ A} = 2 + \dfrac{sin \ A}{cot \ A - cosec \ A}$

Solution:-

First let's simplify LHS for our ease to understand.

$\rm \dfrac{sin \ A}{cosec \ A+ cot \ A}$

• Multiply cosec A - cot A in both numerator and denominator.

$\rm \dashrightarrow\dfrac{sin \ A(cosec \ A - cot \ A)}{(cosec \ A+ cot \ A)(cosec \ A - cot \ A)}$

$\rm \dashrightarrow\dfrac{sin \ A(cosec \ A) - sin \ A( cot \ A) }{cosec^2 \ A - cot^2 \ A} \qquad ...\bigg [ a^2-b^2=(a+b)(a-b)\bigg ]$

We know that,

• sin A = 1 ÷ cosec A

Hence, sin A cosec A = 1

• cot A = cos A/sin A

Hence, sin A cot A = sin × cos A/sin A

→ sin A cot A = cos A

$\rm \dashrightarrow\dfrac{1- cos \ A}{cosec^2 \ A - cot^2 \ A}$

• cosec² A - cot² A = 1

$\rm \dashrightarrow\dfrac{1- cos \ A}{1}$

$\bf \dashrightarrow 1- cos \ A$

Now let's simplify RHS

Our goal is to get the same simplified answer what we got in LHS .

$\rm 2 + \dfrac{sin \ A}{cot \ A - cosec \ A}$

• Multiply cot A + cosec A in both numerator and denominator.

$\rm \dashrightarrow 2 + \dfrac{sin \ A( cot \ A + cosec \ A)}{(cot \ A - cosec \ A)(cot \ A + cosec \ A)}$

$\rm \dashrightarrow 2 + \dfrac{sin \ A( cot \ A) + sin \ A(cosec \ A)}{(cot \ A - cosec \ A)(cot \ A + cosec \ A)}$

As discussed above, same rules will be applied here as well

$\rm \dashrightarrow 2 + \dfrac{cos \ A+ 1}{cot^2\ A - cosec^2 \ A}$

If cosec² A - cot² A = 1, thus

cot² A - cosec² A = -1.

$\rm \dashrightarrow 2 + \dfrac{cos \ A+ 1}{-1}$

$\rm \dashrightarrow 2 + \dfrac{-(cos \ A+ 1)}{1}$

$\rm \dashrightarrow 2 -(cos \ A + 1)$

$\rm \dashrightarrow 2 -cos \ A - 1$

$\bf\dashrightarrow 1-cos \ A$

Hence,

LHS = RHS

Hence, Proved.