Question

(1+tan²A/1+cot²A)=(1-tanA/1-cotA)²=tan²A​

Answer 1

Answer:

Step-by-step explanation:

$\Large{\underline{\underline{\bf{Given:}}}}$

$\sf{\dfrac{1+tan^{2}A }{1+cot^{2}A } =\bigg(\dfrac{1-tan\:A}{1-cot\:A}\bigg) ^{2}=tan^{2}\:A}$

$\Large{\underline{\underline{\bf{To\:Prove:}}}}$

LHS

$\Large{\underline{\underline{\bf{Solution:}}}}$

→ First let us prove

  $\sf{\dfrac{1+tan^{2}A }{1+cot^{2}A } =tan^{2}\:A}$

→ Taking the LHS of the equation

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

→ Using identitities,

  $\sf{\dfrac{1+tan^{2}A }{1+cot^{2}A } =\dfrac{sec^{2}A }{cosec^{2}A }}}$

         $\sf{=\dfrac{1}{cos^{2}A } \div\dfrac{1}{sin^{2}A } }$

         $\sf{=\dfrac{sin^{2}A }{cos^{2}A } }$

         $=\sf{tan^{2} \:A}$

         $\sf{=RHS}$

→ Hence proved.

→ Now we have to prove,

  $\sf{\bigg(\dfrac{1-tan\:A}{1-cot\:A}\bigg) ^{2}=tan^{2}\:A}$

→ Taking the LHS of the equation,

   $\sf{\bigg(\dfrac{1-tan\:A}{1-cot\:A}\bigg) ^{2}}$

→ Simplifying by using identities,

  $\sf{=\left(\dfrac{1-tan\:A}{1-\dfrac{1}{tan\:A} } \right)^{2} }$

→ Cross multiplying,

   $\sf{=\left(\dfrac{1-tan\:A}{\dfrac{tan\:A-1}{tan\:A} } \right)^{2} }$

→ Taking the negative outside,

   $\sf{=\left(\dfrac{-(tan\:A-1)}{\dfrac{tan\:A-1}{tan\:A} } \right)^{2} }$

   $\sf{=\left(-(tan\:A-1)^{2} \div \left(\dfrac{tan\:A-1}{tan\:A} \right)^{2} \right)}$

  $\sf{=\left((tan\:A-1)^{2} \times \left(\dfrac{tan^{2}A }{(tan\:A-1)^{2} } \right) \right)}$

→ Cancelling (tan A - 1)² o both numerator and denominator

  $\sf{=tan^{2}A}$

  $\sf{=RHS}$

→ Hence proved.

→ Therefore,

  $\sf{\dfrac{1+tan^{2}A }{1+cot^{2}A } =\bigg(\dfrac{1-tan\:A}{1-cot\:A}\bigg) ^{2}=tan^{2}\:A}$

$\Large{\underline{\underline{\bf{Identities\:used:}}}}$

1 + tan² A = sec² A

1 + cot² A = cosec² A

1/sec A = cos A

1/cosec A = sin A

sin A/ cos A = tan A

cot A = 1/tan A