Question

it's urgent plzzz tell correct answer plzzzz​

Answer 1

${tan}^{12} \alpha + 1 = {sec}^{2} \alpha \\ ( \frac{5}{12} )^{2} + 1 = {sec}^{2} \alpha \\ \frac{25}{144} + 1 = {sec}^{2} \alpha \\ \frac{25 + 144}{4} = \frac{169}{144} \\ {sec}^{2} \alpha = \frac{169}{144} \\ sec \: \alpha = \sqrt{ \frac{169}{144} }$

= sec alpha = 13/12

Hope it helps...☺

Answer 2

Answer:

tan

12

α+1=sec

2

α

(

12

5

)

2

+1=sec

2

α

144

25

+1=sec

2

α

4

25+144

=

144

169

sec

2

α=

144

169

secα=

144

169

= sec alpha = 13/12

I hope it's helpful

Good morning