Question

(1+ tan2A) +(1+1÷tan2A)=1÷sin2A -sin4A

Answer 1

=

(

1

+

tan

2

A

)

(

1

−

sin

2

A

)

=

sec

2

A

×

cos

2

A

=

1

cos

2

A

×

cos

2

A

=

1

Answer 2

$(1 + {tan}^{2} a)(1 + \frac{1}{ {tan}^{2}a } ) \\ \\ = {sec}^{2} a( \frac{ {tan}^{2} a + 1}{ {tan}^{2} a} ) \\ \\ = \frac{ {sec}^{2} a \times {sec}^{2} a}{ {tan}^{2} a} \\ \\ = \frac{ {sec}^{4}a }{ {tan}^{2} a} \\ \\ = \frac{1}{ {cos}^{2}a {sin}^{2} a} \\ \\ = \frac{1}{ (1 - {sin}^{2} a) {sin}^{2} a } \\ \\ = \frac{1}{ {sin}^{2} a - {sin}^{4} a}$