Question

Sec(2a + 6°) cos(5a + 3º) = 1

Answer 1

Explanation:

sec(2a+6)cos (5a+3)=1

`(cos(5a+3))/(cos(2a+6)) =1\\`

cos(5a+3)=cos(2a+6)

cos(5a+3)-cos(2a+6)=0

`-2sin((5a+3+2a+6))/(2) )sin((5a+3-2a-6))/(2) )=0\\-2sin((7a+9)/(2) )sin((3a-3)/(2) )=0\\either sin ((7a+9)/(2) )=0=sin ~n\pi \\(7a+9)/(2) =n~\pi \\7a+9=2n~\pi \\7a=2n~\pi -9\\a=(2n\pi-9 )/(7) \\where~n~is~an~integer.`

or

`\sin(3a-3)/(2) =0=\sin ~n\pi \\3a-3=2n\pi \\3a=2n\pi+3\\a=(2~n\pi+3 )/(3)`

where n is an integer.