Question

Prove that: tan a/1-cot a + cot a / 1-tan a =1+tan a+cot a

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Step-by-step explanation:

$\large\underline{ \underline{ \sf \maltese{ \: Correct \: Question:- }}}$

$\sf{{Prove \:that:\:}{\dfrac{tanA}{1-cotA}}+{\dfrac{cotA}{1-tanA}}=1+tanA+cotA}$

$\large\underline{ \underline{ \sf \maltese{ \: Proof:- }}}$

LHS:

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

We Know that,

$\boxed{\sf{\red{cot\theta ={\dfrac{1}{tan\theta}}}}}$

Putting Values,

$\sf{{\dfrac{tanA}{1-{\dfrac{1}{tanA}}}}+{\dfrac{{\dfrac{1}{tanA}}}{1-tanA}}}$

$\sf{{\implies}{\dfrac{{tan}^2A}{tanA-1}}+{\dfrac{1}{tanA(tanA-1)}}}$

$\sf{{\implies}{\dfrac{{tan}^3A-1}{tanA(tanA-1)}}}$

$\sf{\green{We\:know\:that:\:}{\boxed{\sf{\green{a^3-b^3=(a-b)(a^2+ab+b^2)}}}}}$

$\sf{{\implies}{\dfrac{(tanA-1)(tan^2A+tanA+1)}{tanA(tanA-1}}}$

$\sf{{\implies}{\dfrac{tan^2A+tanA+1}{tanA}}}$

$\sf{{\implies}{\dfrac{tan^2A}{tanA}}+{\dfrac{tanA}{tanA}}+{\dfrac{1}{tanA}}}$

$\sf{{\implies}{tanA+1+cotA}}$

RHS:

$\sf{1+tanA+cotA}$

$\boxed{\sf{\orange{L.H.S=R.H.S}}}$

Hence, Proved.