Question

T-1
1) If sinA=n-1/n+1
then prove that secA=n+1/2√n​

Answer 1

Answer:

cos A= √(1-sin^2 A)

=√[1-{(n-1)/(n+1)}^2]

=√{4n/(n+1)^2=2√n/(n+1)

sec A =1/cosA=(n+1)/2√n

Answer 2

Step-by-step explanation:

Hope this helps you can neglige the step of finding cosø

instead you can find secø directly however for my sake I've done like this

hope this helps....