Question

prove the following identity-
1/cosecA - cot A - 1/sinA = 1/sinA - 1/cosecA + cotA. I want all answer no copy hrrr​

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, we have,

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

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

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

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

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

From ( 1 ) & ( 2 ) we can conclude that LHS = RHS. Hence, the given identity is proved.

Answer 2

$\mathcal{\huge{\underline{\underline{\red{Question:-}}}}}$

✅ 1/cosecA - cot A - 1/sinA = 1/sinA - 1/cosecA + cotA.

$\mathcal{\huge{\underline{\underline{\orange{Solution:-}}}}}$

Solving LHS :-

1/cosec A - cot A - 1/ sin A

= 1/cosec A - cot A * cosec A + cot A/cosec A + cot A - 1/sin A

 = (cosec A + cot A)/(cosec^2A - cot^2A) - 1/sin A

= (cosec A + cot A)/(cosec^2A - cot^2A) - cosec A

= cosec A + cot A - cosec A

 = cot A.

Solving RHS :-

1/sin A - 1/cosecA+cotA

= cosecA - (cosecA - cotA)/(cosec^2A - cot^2A)

 = cosec A - (cosec A - cot A)

 = cosec A - cosec A + cot A

= cot A.

✨Hence, LHS = RHS.

=> CotA = CotA

☑ So, 1/cosecA - cot A - 1/sinA = 1/sinA - 1/cosecA + cotA.

Hence Proved.✔✔

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