Question

$\rm{\dfrac{1}{cosecA-cotA}-\dfrac{1}{sinA}=\dfrac{1}{sinA}-\dfrac{1}{cosecA+cotA}}$
Prove it ​

Answer 1

To Prove :-

• $\rm{\dfrac{1}{cosecA-cotA}-\dfrac{1}{sinA}=\dfrac{1}{sinA}-\dfrac{1}{cosecA+cotA}}$

Proof :-

Let us start with LHS first. We have,

$:\implies\rm{LHS=\dfrac{1}{cosecA-cotA}-\dfrac{1}{sinA}}$

$:\implies\rm{LHS=\dfrac{1}{(cosecA-cotA)}\times{}\dfrac{(cosecA+cotA)}{(cosecA+cotA)}-\dfrac{1}{sinA}}$

$:\implies\rm{LHS=\dfrac{cosecA+cotA}{cosec^2A-cot^2A}-cosecA}$

$:\implies\rm{LHS=cosecA+cotA-cosecA\quad\quad\:\:[{\because}\:cosec^2A-cot^2A=1]}$

$:\implies\rm{LHS=cotA\longrightarrow{(1)}}$

Now for RHS:-

$:\implies\rm{RHS=\dfrac{1}{sinA}-\dfrac{1}{cosecA+cotA}}$

$:\implies\rm{RHS=\dfrac{1}{sinA}-\dfrac{cosecA-cotA}{(cosecA+cotA)(cosecA-cotA)}}$

$:\implies\rm{RHS=\dfrac{1}{sinA}-\dfrac{cosecA-cotA}{cosec^2A-cot^2A}}$

$:\implies\rm{RHS=cosecA-(cosecA-cotA)\quad\quad[{\because}\:cosec^2A-cot^2A=1]}$

$:\implies\rm{RHS=cotA\longrightarrow{(2)}}$

• Hence,L. H.S =R.H.S

(Proved)

______________________________________

Answer 2

Step-by-step explanation:

Proof :-

Let us start with LHS first. We have,

:\implies\rm{LHS=\dfrac{1}{cosecA-cotA}-\dfrac{1}{sinA}}:⟹LHS=

cosecA−cotA

1

−

sinA

1

:\implies\rm{LHS=\dfrac{1}{(cosecA-cotA)}\times{}\dfrac{(cosecA+cotA)}{(cosecA+cotA)}-\dfrac{1}{sinA}}:⟹LHS=

(cosecA−cotA)

1

×

(cosecA+cotA)

(cosecA+cotA)

−

sinA

1

:\implies\rm{LHS=\dfrac{cosecA+cotA}{cosec^2A-cot^2A}-cosecA}:⟹LHS=

cosec

2

A−cot

2

A

cosecA+cotA

−cosecA

:\implies\rm{LHS=cosecA+cotA-cosecA\quad\quad\:\:[{\because}\:cosec^2A-cot^2A=1]}:⟹LHS=cosecA+cotA−cosecA[∵cosec

2

A−cot

2

A=1]

:\implies\rm{LHS=cotA\longrightarrow{(1)}}:⟹LHS=cotA⟶(1)

Now for RHS:-

:\implies\rm{RHS=\dfrac{1}{sinA}-\dfrac{1}{cosecA+cotA}}:⟹RHS=

sinA

1

−

cosecA+cotA

1

:\implies\rm{RHS=\dfrac{1}{sinA}-\dfrac{cosecA-cotA}{(cosecA+cotA)(cosecA-cotA)}}:⟹RHS=

sinA

1

−

(cosecA+cotA)(cosecA−cotA)

cosecA−cotA

:\implies\rm{RHS=\dfrac{1}{sinA}-\dfrac{cosecA-cotA}{cosec^2A-cot^2A}}:⟹RHS=

sinA

1

−

cosec

2

A−cot

2

A

cosecA−cotA

:\implies\rm{RHS=cosecA-(cosecA-cotA)\quad\quad[{\because}\:cosec^2A-cot^2A=1]}:⟹RHS=cosecA−(cosecA−cotA)[∵cosec

2

A−cot

2

A=1]

:\implies\rm{RHS=cotA\longrightarrow{(2)}}:⟹RHS=cotA⟶(2)

Hence,L. H.S =R.H.S

(Proved)