Question

cotA/1-tanA+tanA/1-cotA=1+secAcosecA prove

Answer 1

LHS=cotA/1-tanA +tanA/1-cotA
cotA=1/tanA
1/tanA(1-tanA) +(tanA)^2/tanA-1
Adding the two fractions ,we get
1-(tanA)^3/tanA(1-tanA)
a^3-b^3=(a-b)(a^2+ab+b^2)
(1-tanA)(1+tanA+(tanA)^2)/tanA(1-tanA)
1-tanA get cancelled
1+(tanA)^2=(secA)^2
tanA+(secA)^2/tanA
splitting
secA×secA/tanA +1
secA =1/cosA,tanA=sinA/cosA,1/sinA=cosecA
substituting,we get
1+secAcosecA

Answer 2

Answer:

LHS

$:\implies \sf \dfrac{\tan(A)}{ 1 - \cos(A) }+ \dfrac{\cot(A)}{ 1 - \tan(A) }$

$:\implies \sf \dfrac{ \dfrac{\sin(A)}{ \cos(A) }}{1 - \dfrac{ \cos(A) }{ \sin(A) }} + \dfrac{ \dfrac{\cos(A)}{ \sin(A) }}{1 - \dfrac{ \sin(A) }{ \cos(A) }}$

$:\implies \sf \dfrac{ \dfrac{\sin(A)}{ \cos(A) }}{\dfrac{ \sin(A) - \cos(A) }{ \sin(A) }} + \dfrac{ \dfrac{\cos(A)}{ \sin(A) }}{ \dfrac{ \cos(A) - \sin(A) }{ \cos(A) }}$

$:\implies \sf \dfrac{\sin(A)}{ \cos(A) } \times \dfrac{\sin(A)}{ \sin(A) - \cos(A) } - \dfrac{\cos(A)}{ \sin(A) } \times \dfrac{\cos(A)}{ \sin(A) - \cos(A) }$

$:\implies \sf \dfrac{\sin^{2} (A)}{ \cos(A) \bigg( \sin(A) - \cos(A) \bigg)} - \dfrac{\cos^{2} (A)}{ \sin(A) \bigg( \sin(A) - \cos(A) \bigg)}$

$:\implies \sf \dfrac{\sin^{3} (A) - \cos^{3} (A)}{ \sin(A)\cos(A) \bigg( \sin(A) - \cos(A) \bigg)}$

$:\implies \sf \dfrac{ \bigg(\sin (A) - \cos (A) \bigg)\bigg(\sin^{2} (A) + \cos^{2} (A) + \sin(A)\cos(A)\bigg)}{ \sin(A)\cos(A) \bigg( \sin(A) - \cos(A) \bigg)}$

$:\implies \sf \dfrac{ 1 + \sin(A).\cos(A)}{ \sin(A)\cos(A) }$

$:\implies \sf \dfrac{1}{ \sin(A) } \times \dfrac{1}{ \cos(A) } + \dfrac{\sin(A)\cos(A)}{ \sin(A)\cos(A) }$

$:\implies \sf \cosec(A) \sec(A) + 1$

$:\implies \textsf{\textbf{ 1 + cosec(A) sec(A) }}$

RHS