Question

Verify :
tanA/1-cotA + cotA/1-tanA =1 + secA cosecA

Answer 1

Step-by-step explanation:

TO PROVE

tanA/1-cotA +  cotA/1-tanA= 1+secA*cosecA

PROOF

LHS:  tanA/1-cotA +cotA/1-tanA

 now, tanA=sinA/cosA , cotA = cosA/sinA

sinA/cosA/1-cosA/sinA+ cosA/sinA/1-sinA/cosA

sinA/cosA/sinA-cosA/sinA+cosA/sinA/cosA-sinA/cosA

sin²A/cosA(sinA-cosA)+cos²A/sinA(cosA-sinA)

sin²A/cosA(sinA-cosA) -cos²A/sinA(sinA-cosA)

sin³A-cos³A/cosA×sinA(sinA-cosA)

using a³-b³=(a-b)(a²+b²+ab)

(sinA-cosA)(sin²A+cos²A+sinA×cosA)/cosA×sinA(sinA-cosA)

[(sinA-cosA)/(sinA-cosA) will be divided and =1]

(sin²A+cos²A+sinA×cosA)/sinA×cosA

using sin²A+cos²A= 1

1/sinA×cosA + sinA×cosA/sinA×cosA

[sinA×cosA/sinA×cosA will be 1]

1/sinA×coAs+1

[sinA= 1/cosecA and cosA = 1/secA]

1/1/cosecA×1/secA+1

cosecA×secA+1

also, 1+secA×cosecA=RHS

HENCE PROVED

LHS=RHS

Answer 2

$\frac{tanA}{1 - tanA} + \frac{cotA}{1 - tanA} \\ \\ = \frac{ \frac{cosA}{sinA} }{ \frac{cosA - sinA}{cosA} } + \frac{ \frac{sinA}{cosA} }{ \frac{sinA - cosA}{sinA} } \\ \\ = \frac{ {cos}^{2}A }{sinA(cosA - sinA)} + \frac{ {sin}^{2}A }{cosA(sinA - cosA)} \\ \\ = \frac{ {cos}^{2} A}{sinA(cosA - sinA)} - \frac{ {sin}^{2}A }{cosA(cosA - sinA)} \\ \\ = \frac{ {cos}^{3}A- {sin}^{3} A }{sinAcosA(cosA - sinA)} \\ \\ = \frac{(cosA - sinA)(1 + sinAcosA)}{sinAcosA(cosA - sinA)} \\ \\ = \frac{1 + sinAcosA}{sinAcosA} \\ \\ = 1 + secAcosecA$