Question

sin A/cot A +cosec A = 2+ sin A/cot A-cosec A

Answer 1

tu
r.h.s
=sinA/cotA+cosecA*cotA- cosecA/cotA-cosecA
=sinA(cotA-cosecA)/cot²A-cosec²A
=sinA* cotA- sinA.cosecA
=sinA.cosA/sinA-1
=cosA-1

l.h.s
=2+ sinA(cotA+ cosecA)/cot²A-cosec²A
=2+ sinA.cotA+ sinA.cosecA
=2+sinA.cosA/sinA+1
=2+cosA-1
=cosA-1
r.h.s=l.h.s

Answer 2

Answer:

To prove:

$2\,+\,\frac{sinA}{cotA\,-\,cosecA}\:=\:\frac{sinA}{cotA\,+\,cosecA}$

proof,

LHS

$=2\,+\,\frac{sinA}{cotA\,-\,cosecA}$

$=2\,+\,\frac{sinA}{cotA\,-\,cosecA}\times\frac{cotA\,+\,cosecA}{cotA\,+\,cosecA}$

$=2\,+\,\frac{sinA(cotA\,+\,cosecA)}{cot^2A\,-\,cosec^2A}$

$=2\,-sinA\times cotA\,+sinA\times\,cosecA$

$=2\,-sinA\times\frac{cosA}{sinA}\,+sinA\times\,\frac{1}{sinA}$

$=2\,-cosA-1$

$=1-cosA$

RHS

$=\:\frac{sinA}{cotA\,+\,cosecA}$

$=\:\frac{sinA}{cotA\,+\,cosecA}\times\frac{cotA\,-\,cosecA}{cotA\,-\,cosecA}$

$=\frac{sinA(cotA\,-\,cosecA)}{cot^2A\,-\,cosec^2A}$

$=-sinA\times cotA\,+sinA\times cosecA$

$=-sinA\times\frac{cosA}{sinA}\,+sinA\times\frac{1}{sinA}$

$=1-cosA$

⇒ LHS = RHS

Hence proved