Question

tan^2tetha-cos^2tetha-1=0​

Answer 1

Answer:

what to do in this?plz explain ur question.

Answer 2

Let theta be A,

$\frac{ { \tan }^{2}theta - 1 }{ { \cos }^{2} theta + 1 } = 0 \\ \\ = \frac{ { \tan}^{2} theta -1 }{ { \cos }^{2} theta + 1} \\ \\ = { \tan}^{2} theta - { \sec }^{2} theta + 1 \\ \\ = { \tan }^{2} theta - (1 + { \tan }^{2} theta) + 1 \\ \\ = { \tan }^{2} theta - 1 - { \tan }^{2} + 1 \\ \\ = 0$

Hence ,Proved