Answer to Question #97936 in Trigonometry for Vishesh bhati

Question

(-tan)^2/(1-cot)^2 = tan^2

Answer 1

Solution:

$$((-tan a)/(1-cot a))^2-〖tan a〗^2=0$$

$$(tan a/(cot a-1))^2-〖tan a〗^2=0$$

$$〖tan a〗^2 (1/(cot a-1)^2 -1)=0$$

$$〖tan a〗^2=0 \,or \, 1/(cot a-1)^2 =1$$

If $〖tan a〗^2=0$ , then tan a=0, so the angle will be a=πn, n $\in$ Z.

If $1/(cot a-1)^2 =1$ , then cot a =2, so the angle will be a=arccot 2+πm, m $\in$ Z

                                 or cot a = 0, but in this situation tan a→∞

Answer:  πn, n $\in$ Z; arccot 2+πm, m $\in$ Z.