Question

tanA/1-cotA+cotA/1-tanA=1+secA.cosecA​

Answer 1

AnswEr :

• To Prove :

$\dfrac{ \tan(A) }{1 - \cot(A) } + \dfrac{ \cot(A) }{1 - \tan(A) } = 1 + \sec(A) \csc(A)$

• Proof :

$\leadsto\dfrac{ \tan(A) }{1 - \cot(A) } + \dfrac{ \cot(A) }{1 - \tan(A) }$

⠀⠀⠀⠀⋆ tan(A) = sin(A) / cos(A)

⠀⠀⠀⠀⋆ cot(A) = cos(A) / sin(A)

$\leadsto\dfrac{ \frac{ \sin(A) }{ \cos(A) } }{1 - \frac{ \cos(A) }{ \sin(A) } } + \dfrac{ \frac{ \cos(A) }{ \sin(A) } }{1 - \frac{ \sin(A) }{ \cos(A) } }$

$\leadsto\dfrac{ \frac{ \sin(A) }{ \cos(A) } }{\frac{ \sin(A) - \cos(A) }{ \sin(A) } } + \dfrac{ \frac{ \cos(A) }{ \sin(A) } }{\frac{ \cos(A) - \sin(A) }{ \cos(A) } }$

$\leadsto\dfrac{ \sin^{2} (A) }{ \cos(A)( \sin(A) - \cos(A)) } + \dfrac{ \cos ^{2} (A) }{ \sin(A)( \cos(A) - \sin(A) )}$

⠀⠀⠀⠀⋆ Taking Negative Common

$\leadsto\dfrac{ \sin^{2} (A) }{ \cos(A)( \sin(A) - \cos(A)) } - \dfrac{ \cos ^{2} (A) }{ \sin(A)( \sin(A) - \cos(A) )}$

$\leadsto \dfrac{1}{( \sin(A) - \cos(A))} \bigg(\dfrac{ \sin^{2} (A) }{ \cos(A) } - \dfrac{ \cos ^{2} (A) }{ \sin(A)} \bigg)$

$\leadsto \dfrac{1}{( \sin(A) - \cos(A))} \bigg(\dfrac{ \sin^{3} (A) - \cos ^{3} (A) }{ \cos(A)\sin(A) } \bigg)$

⠀⠀⠀⠀⋆ (a³ - b³) = (a - b)(a² + b² + ab)

$\leadsto \dfrac{1}{ \cancel{( \sin(A) - \cos(A))}} \times \dfrac{ \cancel{(\sin(A) - \cos(A))}(\sin^{2} (A) + \cos ^{2} (A) + \sin(A) \cos(A)) }{ \cos(A)\sin(A) }$

$\leadsto \dfrac{(\sin^{2} (A) + \cos ^{2} (A) + \sin(A) \cos(A)) }{ \cos(A)\sin(A) }$

⠀⠀⠀⠀⋆ (sin²A + cos²A) = 1

$\leadsto \dfrac{(1 + \sin(A) \cos(A)) }{ \cos(A)\sin(A) }$

$\leadsto \dfrac{1}{ \cos(A)\sin(A) } + \cancel\dfrac{\sin(A) \cos(A) }{ \cos(A)\sin(A) }$

$\leadsto \dfrac{1}{ \cos(A)\sin(A) } + 1$

⠀⠀⠀⠀⋆ 1 / cos(A) = sec(A)

⠀⠀⠀⠀⋆ 1 / sin(A) = cosec(A)

$\leadsto \large1 + \sec(A) \csc(A)$

⠀

$\therefore \boxed{ \rm \dfrac{ \tan(A) }{1 - \cot(A) } + \dfrac{ \cot(A) }{1 - \tan(A) } = 1 + \sec(A) \csc(A) }$