Question

How do I prove that [cosec/(1+sec)]+[(1+sec)/cosec]=2cosec^3 [sec-1]

Answer 1

Given : Cosecθ/(1 + Secθ)  + (1 + Secθ)/Cosecθ  = 2Cosec³θ(Secθ - 1)

To find : to Prove the given

Solution:

Cosecθ/(1 + Secθ)  + (1 + Secθ)/Cosecθ  = 2Cosec³θ(Secθ - 1)

Let say θ = 30°

Cosecθ = 2

secθ = 2/√3

LHS

= 2/(1 + 2/√3)  + ( 1 + 2/√3) /2

= 2√3/(√3 + 2)  + (√3  + 2)/2√3

= 2√3(2 - √3)   + √3(√3  + 2)/6

= ( 12√3(2 - √3)   + 3 + 2√3)/6

= (24√3 - 36 + 3 + 2√3)/6

= (26√3 - 33)/6

≈ 2

RHS =  

2(2)³ (2/√3  - 1)

= 16( 2 - √3 )/√3

= 16√3( 2 - √3 )/3

= 32√3( 2 - √3 )/6

= (64√3 - 96)/6

≈ 2.5

2 ≠ 2.5

LHS ≠ RHS

Hence Data is wrong

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Answer 2

Given:

$\bold{\frac{cosec}{(1+sec)}+\frac{(1+sec)}{cosec}=2\ cosec^3 (\sec-1)}$

To find:

L.H.S=R.H.S

Solution:

In the given equation we assume the value is given in \theta form then question is:

$\frac{\ cosec \theta}{(1+\sec \theta)}+\frac{(1+\sec \theta)}{\ cosec \theta}=2\ cosec^3 \theta(\sec \theta-1)$

Solve L.H.S part:

$\Rightarrow \frac{\ cosec \theta}{(1+\sec \theta)}+\frac{(1+\sec \theta)}{\ cosec \theta}\\\\\Rightarrow \frac{\ cosec^2 \theta+(1+\sec \theta)^2 }{(1+\sec \theta) \ cosec \theta}\\\\\Rightarrow \frac{\ cosec^2 \theta+1+\sec^2 \theta+2\sec \theta }{(1+\sec \theta) \ cosec \theta}\\\\\Rightarrow \frac{\ cosec^2 \theta+1+\sec^2 \theta+2\sec \theta }{(1+\sec \theta) \ cosec \theta}$

Solve R.H.S part:

$\Rightarrow 2\ cosec^3 \theta(\sec \theta-1)\\\\\Rightarrow 2\ \farc{1}{\sin^3 \theta}(\frac{1}{\cos \theta}-1)\\\\\Rightarrow 2 \frac{1}{\sin^3 \theta}(\frac{1-\cos \theta}{\cos \theta})\\\\\Rightarrow \frac{2-2\cos \theta}{\sin^3 \theta \cos \theta}$

if we change the value is sin and cos so, it will prove the R.H.S part that's why we can say that:

L.H.S≠R.H.S