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Answer 1

$\boxed{\underline{\textsf{Formulas :}}}$

$\sf\:1.\:a^{2}+b^{2}=(a+b)^{2}-2ab$

$2.\:tanAcotA =1$

$3.\:a^{2}+b^{2}=(a-b)^{2}+2ab$

$\boxed{\underline{\textsf{Solution :}}}$

$\textsf{Now,}\:tan^{2}A+cot^{2}A=2$

$\to (tanA+cotA)^{2}-2tanAcotA=2$

$\to (tanA+cotA)^{2}-2=2$

$\to (tanA+cotA)^{2}=2+2$

$\to (tanA+cotA)^{2}=4$

$\to \boxed{tanA+cotA=\pm 2}$

$\textsf{Again,}\:tan^{2}A+cot^{2}A=2$

$\to (tanA-cotA)^{2}+2tanAcotA=2$

$\to (tanA-cotA)^{2}+2=2$

$\to (tanA-cotA)^{2}=2-2$

$\to (tanA-cotA)^{2}=0$

$\to \boxed{tanA-cotA= 0}$

Answer 2

Dude

Answer is in the attachment

It depends on formula

Used formulas:

A^2 +B^2 =(A+B) ^2 - 2AB

(A+B) ^2=(A-B)^2 +4AB