Question

PROVE THAT : sinA / cotA +cosecA = 2+ sinA / cotA - cosecA

Answer 1

LHS
=sinA/cotA+cosecA
=sinA/(cosA/sinA+1/sinA)
=sinA/{(cosA+1)/sinA}
=sin²A/(1+cosA)
=(1-cos²A)/(1+cosA)
=(1+cosA)(1-cosA)/(1+cosA)
=1-cosA
RHS
=2+sinA/cotA-cosecA
=2+sinA/(cosA/sinA-1/sinA)
=2+sinA/{(cosA-1)/sinA}
=2+sin²A/{-(1-cosA)}
=2-{(1-cos²A)/(1-cosA)}
=2-{(1+cosA)(1-cosA)/(1-cosA)}
=2-(1+cosA)
=2-1-cosA
=1-cosA
∴, LHS=RHS

Answer 2

Step-by-step explanation:

Given that:

$\frac{sinA}{cotA + cosec A} = 2 + \frac{sin A}{cotA - cosec A}$

Solution:

Take LHS

Multiply and divide by

$cotA - cosec A$

$\frac{sin A}{cotA + cosec A} \times \frac{cotA - cosec A}{cotA - cosecA} \\ \\ = \frac{sin A(cotA - cosec A)}{ {cot}^{2} A - {cosec}^{2} A} \\ \\ \because {cosec}^{2}A - {cot}^{2} A = 1 \\ \\ = \frac{sin A(cosecA - cot A)}{1} \\ \\ = sin A\bigg( \frac{1}{sin A} - \frac{cos A}{sin A} \bigg)$

$= sin A\bigg( \frac{1 - cos A}{sinA}\bigg ) \\ \\ = 1 - cos A \: \: \: \: ...eq1$

Now take RHS

$2 + \frac{sin A}{cotA - cosec A} \\ \\ = 2 + \frac{sin A}{cotA - cosec A} \times \frac{cotA + cosec A}{cotA + cosec A} \\ \\ = 2 - \frac{sin A(cotA + cosec A)}{ {cosec}^{2} A - {cot}^{2} A } \\ \\ = 2 - sin A(cotA + cosec A) \\ \\ = 2 - sin A\bigg(\frac{cos A}{sinA} + \frac{1}{sin A} \bigg) \\ \\ = 2 - \frac{sin A(1 + cos A)}{sin A} \\ \\ = 2 - 1 - cos A \\ \\ = 1 - cos A \: \: \: \: ...eq2$

From Eq1 and eq2

it is proved that

$\frac{sinA}{cotA + cosec A} = 2 + \frac{sin A}{cotA - cosec A}$

Hope it helps you.