Question

show that cos15°-sin15°/cos15°+sin15° = 1/√3

Answer 1

Hi ,

LHS = (cos15- sin15)/(cos15+sin15)

= [(cos15-sin15)(cos 15-sin15)]/[(cos15+sin15)(cos 15-sin15)]

= ( Cos15-sin15 )²/( cos²15 - sin² 15 )

= [cos²15+sin² 15-2sin15cos15]/cos(2×15)

= [ 1 - sin ( 2×15 )]/ cos30

= ( 1 - sin30 ) / cos30

= ( 1 - 1/2 ) / ( √3/2 )

= ( 1/2 ) / ( √3/2 )

= 1/√3

= RHS

I hope this helps you.

: )

Answer 2

On LHS we have = (cos15- sin15)/(cos15+sin15)

= [(cos15-sin15)(cos 15-sin15)]/[(cos15+sin15)(cos 15-sin15)]

= ( Cos15-sin15 )²/( cos²15 - sin² 15 )

= [cos²15+sin² 15-2sin15cos15]/cos(2×15)

= [ 1 - sin ( 2×15 )]/ cos30

= ( 1 - sin30 ) / cos30

= ( 1 - 1/2 ) / ( √3/2 )

= ( 1/2 ) / ( √3/2 )

= 1/√3

= RHS

Hence proved.