Question

Prove : tan A/( 1-cot A) + cot A / (1-tan A) = sec A.cosec A +1
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Answer 1

To Prove

tanA/(1 - cotA) + cotA/(1 - tanA) = secAcosecA + 1

Proof

Taking LHS,

We can write tanA as secA/cosecA and cotA as cosecA/secA. Hence the expression becomes

$\sf\frac{tanA}{1 - cotA} + \frac{cotA}{1 - tanA}$

$\sf\frac{\frac{secA}{cosecA}}{1 - \frac{cosecA}{secA}} + \frac{\frac{cosecA}{secA}}{1 - \frac{secA}{cosecA}}$

$\sf\frac{\frac{secA}{cosecA}}{\frac{secA -cosecA}{secA}} + \frac{\frac{cosecA}{secA}}{\frac{cosecA-secA}{cosecA}}$

$\sf\frac{sec^2A}{cosecA(secA - cosecA)} + \frac{cosec^2A}{secA(cosecA - secA)}$

Taking out minus sign from denominator of second term, we get

$\sf\frac{sec^2A}{cosecA(secA - cosecA)} - \frac{cosec^2A}{secA(secA - cosecA)}$

Now multiply with secA in both numerator and denominator on first term and with cosecA in both numerator and denominator on the second to get equal denominators.

$\sf\frac{sec^3A}{cosecAsecA(secA - cosecA)} - \frac{cosec^3A}{cosecAsecA(secA - cosecA)}$

Split using a³ - b³ = (a - b)(a² + b² + ab)

$\sf\frac{(secA - cosecA)(sec^2A + cosec^2A + secAcosecA)}{secAcosecA(secA - cosecA)}$

Cancel out (secA - cosecA)

$\sf\frac{sec^2A + cosec^2A + secAcosecA}{secAcosecA}$

Now,

sec²A + cosec²A = 1/cos²A + 1/sin²A

→ sec²A + cosec²A = (sin²A + cos²A)/cos²Asin²A

→ sec²A + cosec²A = 1/sin²Acos²A

→ sec²A + cosec²A = sec²cosec²A

Put this value in expression and it becomes

$\sf\frac{sec^2Acosec^2A + secAcosecA}{secAcosecA}$

Take out secAcosecA common in numerator

$\sf\frac{secAcosecA(secAcosecA + 1)}{secAcosecA}$

Cancel out secAcosecA

= $\sf secAcosecA + 1$

Hence proved.

Answer 2

Question :--- Prove that tan A/( 1-cot A) + cot A / (1-tan A) = sec A.cosec A +1

Formula used :---

• CotA = 1/TanA
• (a³ - b³) = (a-b)(a²+b²+ab)
• 1 + sec²A = tan²A
• SecA = 1/cosA
• cosecA = 1/sinA
• TanA = sinA/cosA

Solution :-----

Solving LHS we get,

tanA/(1-cotA) +cotA/(1-tanA)

Putting CotA = 1/tanA we get,

→ tanA/(1–1/tanA) +cotA/(1-tanA)

Taking LCM and Taking Denominator on Multiply we get,

→ tan²A/(tanA-1) +cotA/(1-tanA)

Taking (-1) common From First part now,

→ (-tan²A)/(1-tanA) +cotA/(1-tanA)

→ (-tan²A+cotA)/(1-tanA)

again putting cotA = 1/tanA now,

→ (-tan²A+1/tanA)/(1-tanA)

Taking LCM again

→ (-tan³A+1)/tanA(1-tanA)

→ (1-tan³A)/tanA(1-tanA)

using (a³ - b³) = (a-b)(a²+b²+ab) now,

→ (1-tanA)(1+tanA+tan²A)/tanA(1-tanA)

(1-tanA) will be cancel From both sides ,

→ (1+tanA+tan²A)/tanA

→ [(1+tan²A)+tanA]/tanA

Using (1+tan²A) = sec²A now,

→ (sec²A+tanA)/tanA

→ sec²A/tanA + (tanA/tanA)

Putting secA = 1/cosA now,

→ 1/cos²AtanA + 1

Putting TanA = sinA/cosA now,

→ cosA/cos²AsinA + 1

→ 1/cosAsinA + 1

using 1/cosA = secA and 1/sinA = cosecA now,

→ secAcosecA + 1 = RHS .

✪✪ Hence Proved ✪✪