Question

please answer the question as fast as possible I will mark you as Brainliest... ​

Answer 1

Step-by-step explanation:

LHS:

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

put tanA=sinA/cosA and cotA=cosA/sinA

= sin^2A/cosA(sinA-cosA) + cos^2A/sinA(cosA-sinA)

=sin^2A/cosA(sinA-cosA) - cos^2A/sinA(sinA-cosA)

Taking LCM

= (sin^3A-cos^3A)/sinAcosA(sinA-cosA)

We know that a^3-b^3=(a-b)(a^2+ab+b^2)

=(sinA-cosA)(sin^2A+cos^2A+sinAcosA)/sinAcosA(sinA-cosA)

(sinA-cosA) of the numerator and the denominator cancel each other.

Also, sin^2A + cos^2A = 1

= (1+sinAcosA)/sinAcosA

= 1/sinAcosA + sinAcosA/sinAcosA

=cosecAsecA+1

RHS:

1+tanA+cotA

=1+sinA/cosA+cosA/sinA

Taking LCM

=(sinAcosA+sin^2A+cos^2A)/sinAcosA

=(sinAcosA+1)/sinAcosA

=1+cosecAsecA

=cosecAsecA+1

           LHS=RHS

Hence proved

Hope this answer helps you:)