Question

pls answer this fast I will mark as brainlist​

Answer 1

Answer:

PQ = 13, QR = 5

=> PR = √(13²-5²) = √(169-25) = √144 = 12

$sin \alpha \: = \: 5 \div 13 \\ cos \alpha \: = 12 \div 13 \\ sintheta \: = 12 \div 13 \\ costheta \: = 5 \div 13$

1) L.H.S = sin²alpha + cos²alpha = 25/169 + 144/169

= 169/169 = 1

R.H.S = sin²theta + cos²theta = 144/169 + 25/169

= 169/169 = 1

since, L.H.S = R.H.S

the first equation proved

$sec \alpha = 13 \div 12 \\ tan \alpha = 5 \div 12 \\ sec(theta) = 13 \div 5 \\ tan(theta) = 12 \div 5$

2) L.H.S = sec²alpha - tan²alpha = 169/144 - 25/144

= 144/144 = 1

R.H.S = sec²theta - tan²theta = 169/25 - 144/25

= 25/25 = 1

since, L.H.S = R.H.S

the 2nd equation is true