Question

prove that tan a by 1 minus cot A + cot a by 1 minus Tan a equal to 1 + tan A + cot a equal to 1 + sec a + cosec A​

Answer 1

Correct Question

Prove that tanA/(1 - cotA) + cotA/(1 - tanA) = 1 + tanA + cotA = 1 + secA.cosecA

To Prove

tanA/(1 - cotA) + cotA/(1 - tanA) = 1 + tanA + cotA = 1 + secA.cosecA

$\rule{200}2$

Proof

Taking L.H.S.

⇒ tanA/(1 - cotA) + cotA/(1 - tanA)

We know that, tanA = sinA/cosA and cotA = cosA/sinA

⇒ $\sf{\frac{ \frac{sinA}{cosA} }{1 - \frac{cosA}{sinA} } + \frac{ \frac{cosA}{sinA} }{ 1 - \frac{sinA}{cosA} }}$

⇒ $\sf{\frac{ \frac{sinA}{cosA} }{\frac{sinA-cosA}{sinA} } + \frac{ \frac{cosA}{sinA} }{\frac{cosA-sinA}{cosA} }}$

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

⇒ $\sf{\frac{sin^3A-cos^3A}{cosA\:sinA(sinA-cosA)}}$

Used identity: (a - b)³= (a - b) (a² + b²+ ab)

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

⇒ $\sf{\frac{sin^2A+cos^2A+sinA.cosA}{cosA.sinA}}$

Also, sin²A + cos²A = 1

⇒ (1 + cosA.sinA)/(cosA.sinA)

⇒ 1/cosA.sinA + cosA.sinA/cosA.sinA

Now, 1/cosA = secA and 1/sinA = cosecA

⇒ secA.cosecA + 1

Now take, 1+ tanA + cotA

⇒ 1+ sinA/cosA + cosA/sinA

⇒ $\sf{\frac{sin^2A+cos^2A+sinA.cosA}{cosA.sinA}}$

⇒ (1 + cosA.sinA)/(cosA.sinA)

⇒ 1/cosA.sinA + cosA.sinA/cosA.sinA

⇒ secA.cosecA + 1

Hence, proved

Answer 2

||✪✪ QUESTION ✪✪||

Prove That, (TanA/(1 - CotA) + cotA/(1 - TanA) = 1 + TanA + cotA = 1 + SecA * cosecA ?

|| ✰✰ ANSWER ✰✰ ||

→ (TanA/(1 - CotA) + cotA/(1 - TanA)

Putting CotA = (1/TanA) we get,

→ (TanA/( 1 - (1/TanA)) + (1/TanA) / (1 - TanA)

→ Tan²A/(TanA - 1) + 1/TanA(1 - TanA)

Taking (-1) common now,

→ Tan²A/(TanA - 1) - 1/TanA(TanA - 1)

Taking LCM Now,

→ (Tan³A - 1) /TanA(TanA - 1)

→ (Tan³A - 1³) /TanA(TanA - 1)

Using (a³ - b³) = (a - b)(a² + b² + ab) in Numerator now,

→ [ (TanA - 1)(Tan²A + TanA + 1) ] / (TanA(TanA - 1))

(TanA - 1) will be cancel From both sides now,

we get,

______________________________

→ (Tan²A + TanA + 1) / TanA ----------- Equation (1) .

Assume This As Equation (1) now, we will prove Our RHS part With This now :-

→ (Tan²A + TanA + 1) / TanA

we know That, (Tan²A + 1) = sec²A , so, using this we get,

→ (sec²A + TanA) / TanA

→ (sec²A/TanA) + 1

Now, putting sec²A = (1/cos²A) and TanA = (sinA/cosA),

→ [ (1/cos²A) / (sinA/cosA) ] + 1

→ (1/sinA * cosA) + 1

Now, we know That, (1/sinA) = cosecA and (1/cosA) = secA , So, Finally we get,

→ 1 + secA * cosecA . ✪✪ Hence Proved ✪✪

_____________________________

Now, Again, Solving From Equation (1) , we get :-

→ (Tan²A + TanA + 1) / TanA

→ (Tan²A/TanA) + (TanA/TanA) + (1/TanA)

→ TanA + 1 + (1/TanA)

As we know , (1/TanA) = cotA, So,

→ 1 + TanA + cotA . ✪✪ Hence Proved ✪✪

_____________________________