Question

prove that


cotA-cos A /cot A +cos A =  cosec A-1/cosec A +1


this is the question in the image explain the (1)marked step

Answer 1

GIVEN :

The trignometric equation is $\frac{cotA-cosA}{cotA-cosA}=\frac{cosecA-1}{cosecA+1}$

TO PROVE :

The given equality  $\frac{cotA-cosA}{cotA-cosA}=\frac{cosecA-1}{cosecA+1}$ is true.

SOLUTION :

Given that  $\frac{cotA-cosA}{cotA+cosA}=\frac{cosecA-1}{cosecA+1}$

Now we have to prove that LHS=RHS

Now taking the LHS

$\frac{cotA-cosA}{cotA+cosA}$

Multiply and divide the above expression by denominator's conjugate $cotA-cosA$

$=\frac{cotA-cosA}{cotA+cosA}\times \frac{cotA-cosA}{cotA-cosA}$

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

By using the algebraic identities :

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

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

$=\frac{cot^2A-2cotAcosA+cos^2A}{cot^2A-cos^2A}$

By using the trignometric property :

$cotx=\frac{cosx}{sinx}$

$=\frac{\frac{cos^2A}{sin^2A}-2(\frac{cosA}{sinA}).cosA+cos^2A}{\frac{cos^2A}{sin^2A}-cos^2A}$

$=\frac{\frac{cos^2A}{sin^2A}-2(\frac{cos^2A}{sinA})+cos^2A}{\frac{cos^2A}{sin^2A}-cos^2A}$

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

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

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

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

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

By using the algebraic identity :

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

$=\frac{(1-sin^2A)^2}{1^2-sin^2A}$

By using the algebraic identity :

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

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

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

By using the trignometric property :

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

$=\frac{1-\frac{1}{cosecA}}{1+\frac{1}{cosecA}}$

$=\frac{\frac{cosecA-1}{cosecA}}{\frac{cosecA+1}{cosecA}}$

$\frac{cosecA-1}{cosecA+1}$=RHS

∴ LHS=RHS

∴  $\frac{cotA-cosA}{cotA+cosA}=\frac{cosecA-1}{cosecA+1}$

Hence proved.

Answer 2

Answer:

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