Question

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

Answer 1

Step-by-step explanation:

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Answer 2

Step-by-step explanation:

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

Consider: (1+tan²A/1+cot²A)

[1+tan²A=sec²A. and. 1+cot²A=cosec²A.]

=> [sec²A/cosec²A]

[sec²A=1/cos²A. and cosec²A=1/sin²A]

=> [1/cos²A/1/sin²A]

=> [sin²A/cos²]

=> tan²A.

Now consider:(1-tanA/1-cotA)²

$= > { (\frac{1 - \frac{ \sin(a) }{ \cos(a) } }{1 - \frac{ \cos(a) }{ \sin(a) } }) }^{2} = > {( \frac{ \frac{ \cos(a) - \sin(a) }{ \cos(a) ) } }{ \frac{ \sin(a) - \cos(a) }{ \sin(a) } } )}^{2}$

$= > {( \frac{ - ( \sin(a) - \cos(a) ) \times \sin(a) }{( \sin(a) - \cos(a) ) \times \cos(a) } )}^{2}$

[(sinA-cosA) cancels]

=>(-sinA/cosA)²

=>(-tanA)²

=>tan²A.

All are equal.

Hence proved.