Question

tan A cot A
Prove that
1-cot A 1-tan A
1+tan A+ cot A. CB​

Answer 1

Answer:

I think your question is, prove that,

tanA/1-cotA + CotA/1-tanA = 1+tanA+cotA, if so, see the below proof.

Proof:

take LHS

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

= $\frac{tanA}{1-\frac{1}{tanA} }$ + $\frac{\frac{1}{tanA} }{1-tanA}$

= $\frac{tanA}{\frac{tanA-1}{tanA} }$ + $\frac{1 }{tanA(1-tanA)}$

= $\frac{tan^2A}{tanA-1 }$ + $\frac{1 }{tanA(1-tanA)}$

= $-\frac{tan^2A}{1-tanA}$ + $\frac{1 }{tanA(1-tanA)}$

=  $\frac{-tan^3A+1 }{tanA(1-tanA)}$ = $\frac{1^3 - tan^3A }{tanA(1-tanA)}$

= $\frac{(1 - tanA)(1^2+tan^2A+(1)(tanA) }{tanA(1-tanA)}$    (   ∵ a³-b³ = (a-b)(a²+b²+ab)  )

= $\frac{1+tan^2A+tanA }{tanA}$

= $\frac{1}{tanA}$ + $\frac{tan^2A}{tanA}$ + $\frac{tanA}{tanA}$

= cotA + tanA + 1

= 1+tanA+cotA

= RHS

=> LHS = RHS

Hence proved