Question

Prove that
$\bf{ \frac{cosA - sinA+1}{cosA + sinA - 1} } = cosecA + cotA$
​

Answer 1

$\huge\star\underline\mathfrak\green{Answer:-}$

cosA-sinA+1/cosA+sinA-1

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

=(cos²A-cosAsinA+cosA+cosAsinA-sin²A+sinA+cosA-sinA+1)/{(cosA+sinA)²-(1)²}

=(cos²A-sin²A+2cosA+1)/(cos²A+2cosAsinA+sin²A-1)

={cos²A+2cosA+(1-sin²A)}/(1+2cosAsinA-1) [∵, sin²A+cos²A=1]

=(cos²A+2cosA+cos²A)/2cosAsinA

=(2cos²A+2cosA)/2cosAsinA

=2cosA(cosA+1)/2cosAsinA

=(cosA+1)/sinA

=cosA/sinA+1/sinA

=cotA+cosecA

=cosecA+cotA (Proved)

Answer 2

Given Identity:-

• $\sf{\dfrac{CosA - SinA + 1}{CosA + SinA - 1} = CosecA + CotA}$

To Proof:-

• LHS = RHS

Solution:-

Taking LHS,

$\sf{\dfrac{CosA - SinA + 1}{CosA + SinA - 1}}$

Dividing both numerator and denominator by SinA

= $\sf{\dfrac{\dfrac{CosA}{SinA} - \dfrac{SinA}{SinA} + \dfrac{1}{SinA}}{\dfrac{CosA}{SinA} + \dfrac{SinA}{SinA} - \dfrac{1}{SinA}}}$

Since, $\sf{\dfrac{CosA}{SinA} = CotA}$

And, $\sf{\dfrac{1}{SinA} = CosecA}$

Also, $\sf{\dfrac{SinA}{SinA} = 1}$

= $\sf{\dfrac{CotA - 1 + CosecA}{CotA + 1 - CosecA}}$

= $\sf{\dfrac{CotA + CosecA - 1}{CotA - CosecA + 1}}$

Now,

Since, $\sf{1+Cot^2A = Cosec^2A}$

Therefore,

$\sf{1 = Cosec^2A - Cot^2A}$

Putting 1 = Cosec²A - Cot²A in the numerator

$\sf{\dfrac{CotA + CosecA - (Cosec^2A - Cot^2A)}{CotA - CosecA + 1}}$

Now applying another identity i.e.,

a² - b² = (a + b) (a - b)

= $\sf{\dfrac{CotA + CosecA - [(CosecA + CotA)(CosecA - CotA)]}{CotA - CosecA + 1}}$

Taking (CotA + CosecA) as common,

= $\sf{\dfrac{(CotA + CosecA)[1 - (CosecA - CotA)]}{CotA - CosecA + 1}}$

= $\sf{\dfrac{(CotA + CosecA)(1-CosecA + CotA)}{CotA - CosecA + 1}}$

= $\sf{\dfrac{(CotA + CosecA)(CotA - CosecA + 1)}{CotA - CosecA + 1}}$

= $\sf{\dfrac{(CotA + CosecA)(\cancel{CotA - CosecA + 1)}}{\cancel{CotA-CosecA + 1}}}$

= $\sf{CotA + CosecA}$

Hence LHS = RHS (Proved)

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Identities Used:-

• 1 + Cot²A = Cosec²A
• a² - b² = (a + b)(a - b)

• $\sf{\dfrac{CosA}{SinA} = CotA}$

• $\sf{\dfrac{1}{SinA} = CosecA}$

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Explore More!!

Other identities:-

• Sin²A + Cos²A = 1
• 1 + Tan²A = Sec²A

• $\sf{\dfrac{SinA}{CosA} = TanA}$

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