Question

1minus tan square thetha by cot square theatha minus1 = tan square theatha

Answer 1

Step-by-step explanation:

let0=theta

tan^20=1÷cot^2 0

take lcm

then cross multiply

then u will get ur answer

Answer 2

LHS=RHS

Hence proved

$\frac{1 - \tan ^{2} Ѳ }{cot Ѳ ^{2} - 1 } = \tan ^{2} Ѳ \\ LHS \\ \frac{1 - \tan ^{2} Ѳ } { \: \frac{1}{ \tan ^{2} Ѳ } - 1} \: \\\frac{1 - \tan ^{2} Ѳ } { \: \frac{1 - \tan^{2} Ѳ }{ \tan ^{2} Ѳ } } \\ 1 - \tan ^{2} Ѳ \times \frac{ \tan ^{2} Ѳ }{1 - \tan ^{2} Ѳ} \\ \tan ^{2} Ѳ =RHS \\ </p><p>Hence \: proved$