Question

1+sin theta upon cos theta+cos theta upon 1+sin theta=2sec

Answer 1

Hi ,

Here I used A instead of theta.

LHS = ( 1+ sinA )/ cosA + cosA/( 1 + sinA )

= [ ( 1 + sinA )² + cos² A ]/[ cosA(1+sinA ) ]

= ( 1 + sin² A + 2sinA + cos² A )/[ cosA (1+sinA )]

= [ 1 + 2sinA + ( sin² A + cos² A)]/[cosA (1+sinA)]

= [ 1+2sinA + 1 ]/[ cosA ( 1 + sinA ) ]

= ( 2 + 2sinA )/[ cosA( 1 + sinA) ]

= 2( 1 + sinA )/[ cosA( 1 + sinA ) ]

= 2/cosA

= 2secA

= RHS

I hope this helps you.

: )

Answer 2

Hi,

Please see the attached file!


Thanks