Question

( 1/ cosA +sin A - 1 ) + (1/cos A + sin A +1 ) = cosec A + sec A​

Answer 1

GIVEN :

The equation is $\frac{1}{cosA+sinA-1}+\frac{1}{cosA+sinA+1}=cosecA+secA$

TO PROVE :

The given equation is true.

SOLUTION :

Given equation is $\frac{1}{cosA+sinA-1}+\frac{1}{cosA+sinA+1}=cosecA+secA$

We have to prove that the given equation is true.

That is prove that LHS = RHS

Taking LHS $\frac{1}{cosA+sinA-1}+\frac{1}{cosA+sinA+1}\hfill (1)$

Now $\frac{1}{(cosA+sinA)-1}$

Multiply and divide by denominator's conjugate,

$\frac{1}{(cosA+sinA)-1}\times \frac{(cosA+sinA)+1}{(cosA+sinA)+1}$

$=\frac{(cosA+sinA)+1}{(cosA+sinA)^2-1^2}$

By using the algebraic identity :

$(a+b)(a-b)=a^2-b^2$

$=\frac{cosA+sinA+1}{cos^2A+2cosA sinA+sin^2A-1}$

By using the trignometric identity :

$sin^2x+cos^2x=1$

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

$\frac{1}{(cosA+sinA)-1}=\frac{cosA+sinA+1}{2cosA sinA}\hfill (2)$

Now $\frac{1}{(cosA+sinA)+1}$

Multiply and divide by denominator's conjugate,

$\frac{1}{(cosA+sinA)+1}\times \frac{(cosA+sinA)-1}{(cosA+sinA)-1}$

$=\frac{(cosA+sinA)-1}{(cosA+sinA)^2-1^2}$

By using the algebraic identity :

$(a+b)(a-b)=a^2-b^2$

$=\frac{cosA+sinA-1}{cos^2A+2cosA sinA+sin^2A-1}$

By using the trignometric identity :

$sin^2x+cos^2x=1$

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

$\frac{1}{(cosA+sinA)+1}=\frac{cosA+sinA-1}{2cosA sinA}\hfill (3)$

Substituting the equations (2) and (3) we get,

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

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

$=\frac{2cosA+2sinA}{2cosA sinA}$

$=\frac{2(cosA+sinA)}{2cosA sinA}$

$=\frac{cosA+sinA}{cosA sinA}$

$=\frac{cosA}{cosA sinA}+\frac{sinA}{cosA sinA}$

$=\frac{1}{sinA}+\frac{1}{cosA}$

By using the trignometric identities :

$\frac{1}{cosx}=secx$

$\frac{1}{sinx}=cosecx$

$=cosecA+secA$ =RHS

∴ LHS = RHS

⇒ $\frac{1}{cosA+sinA-1}+\frac{1}{cosA+sinA+1}=cosecA+secA$

Hence proved

Answer 2

Answer:

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