Question

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

Answer 1

hello

your answer....↓↓

⇒ LHS =  

sinA/(cotA+cosecA)

= (sinA)(cosecA-cotA)/(cosecA+cotA)(cosecA-cotA)

= (1-cosA)/(cosec2A-cot2A)

= (1-cosA)/1 = 1-cosA

⇒ RHS =  

2+(sinA)/(cotA-cosecA)

= 2+(sinA)(cotA+cosecA)/(cotA-cosecA)(cotA+cosecA)

= 2+(cosA+1)/(cot2A-cosec2A)

= 2+(cosA+1)/(-1)

= 2-cosA-1

= 1-cosA

LHS=RHS=1-cosA

regards

:)

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

$cotA - cosec A$

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

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

Now take RHS

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

From Eq1 and eq2

it is proved that

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

Hope it helps you