Question

please solve this question... ​

Answer 1

Answer:

Given:

The equation is,

CosA-SinA+1/CosA+SinA-1=CosecA+CotA

Formula used:

→ Cosec²A-cot²A=1

→ a²-b²=(a+b)(a-b)

Solution:

First Take LHS,

→ CosA-SinA+1/CosA+SinA-1

Divide each term by 'SinA'.

→ cosA/sinA-1+1/SinA/cosA/SinA+1-1/sinA

→ CotA-1+CosecA/cotA+1-cosecA

As we know that,

Cosec²A-cot²A=1

→ CotA+CosecA-(cosec²A-cot²A)/ CotA-CosecA+1

→ CotA+CosecA-((CosecA+cotA)(CosecA-cotA))/ cotA-CosecA+1

→ CotA+CosecA (1-(CosecA-cotA))/cotA-CosecA+1

→ cotA+CosecA (1-cosecA+cotA)/cotA-CosecA+1

→ CotA+CosecA

So, LHS=RHS

Answer 2

Answer:

$\Large{\underline{\underline{\bf{Given:}}}}$

$\dfrac{cos\:A-sin\:A\:+1}{cos\:A+sin\:A-1} = cosec\:A+cot\:A$

$\Large{\underline{\underline{\bf{To\:Prove:}}}}$

LHS = RHS

$\Large{\underline{\underline{\bf{Proof:}}}}$

→ Divide throughout by sin A on LHS side

$\dfrac{\dfrac{cos\:A}{sin\:A}-\dfrac{sin\:A}{sin\:A}+\dfrac{1}{sin\:A} }{\dfrac{cos\:A}{sin\:A}+\dfrac{sin\:A}{sin\:A} -\dfrac{1}{sin\:A} }$

$\dfrac{cot\:A-1+cosec\:A}{cot\:A+1-cosec\:A}$

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

$\dfrac{cot\:A+cosec\:A-(cosec\:A-cot\:A)(cosec\:A+cot\:A)}{1-cosec\:A+cot\:A}$

Taking cot A + cosec A common from the numerator

$\dfrac{cot\:A+cosec\:A(1-cosec\:A+cot\:A)}{1-cosec\:A+cot\:A}$

Cancelling 1 - cosec A + cot A on both numerator and denominator

$= cosec\:A+cot\:A = RHS$

Hence proved.

$\Large{\underline{\underline{\bf{Identities\:used:}}}}$

➳ $\dfrac{cos\:A}{sin\:A} = cot\:A$

➳ $\dfrac{1}{sin\:A}=cosec\:A$

➳ $cosec^{2} \:A-cot^{2}\:A=1$

➳ $(a^{2} -b^{2} )= (a +b)\times (a-b)$