Question

prove that :
(cosecA- cotA)² = 1-cosA/1+cosA ​

Answer 1

Step-by-step explanation:

(cosecA-cotA)²

=(1-cosA/sinA)²

=(1-cosA)²/sin²A

=(1-cosA)²/(1-cosA)(1+cosA)

=(1-cosA)/(1+cosA)

Answer 2

Given LHS :

• $\sf(cosec \: a - cot \: a) ^{2}$

Given RHS :

• $\sf \dfrac{1 - cos \: a}{1 +cos \: a}$

To prove :

• $\boxed{\bold{(cosec\:A-\:cot\:A)^2}=\dfrac{1-cos\:A}{1+cos A}}$

Proof :

Taking LHS,

$\sf{\implies{(cosec\:A-\:cot\:A)^2}}$

$\sf{\implies{\Big(\dfrac{1}{sin\:A}-\:\dfrac{cos\:A}{sin\:A}\Big)^2}}$

$\bold{\big[\because\:Cosec\:A\:=\:\dfrac{1}{sin\:A}\:\:;\:\:cot\:A=\:\dfrac{cos\:A}{sin\:A}\big]}$

$\sf{\implies{\Big(\dfrac{1-cos\:A}{sin\:A}\Big)^2}}$

$\sf{\implies{\dfrac{(1-cos\:A)^2}{sin^2\:A}}}$

$\sf{\implies{\dfrac{(1-cos^2\:A)}{1^2-cos^2\:A}}}$

$\bold{\big[\because\:Sin^2\:A=\:1\:-\:cos^2\:A\big]}$

$\sf{\implies{\dfrac{(1-cos^2A)}{(1+cosA)(1-cosA)}}}$

$\bold{\big[\because\:a^2-b^2=(a+b) (a-b)\big]}$

$\sf{\implies{\dfrac{1-cos\:A}{1+cos\:A}}}$

$\boxed{\bold{(cosec\:A-cot\:A)^2=\dfrac{1-cos\:A}{1+cos\:A}}}$