(sin 42 degree + cos 48 degree) (sin 42 degree - Cos 48 degree)
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1
(sin42° + cos48°) (sin42° - cos48°)
=(sin42°)²-(cos48°)² [(a+b)(a-b)=a²-b²]
=(sin42°)²-(cos(90-42))²
=sin²42°-sin²42°
=0
=(sin42°)²-(cos48°)² [(a+b)(a-b)=a²-b²]
=(sin42°)²-(cos(90-42))²
=sin²42°-sin²42°
=0
Answered by
1
Heya !!!
Sin (90- Theta) = Cos theta
(Sin 42° + Cos48°) (Sin42° -Cos48°)
{Sin (90- 48°) + Cos 48° } { Sin(90-48°) - cos48*}
=> (Cos48° + Cos48° ) ( Cos48° - Cos48°)
=> (Cos²48°) - (Cos²48)
=> 0
HOPE IT WILL HELP YOU...... :-)
Sin (90- Theta) = Cos theta
(Sin 42° + Cos48°) (Sin42° -Cos48°)
{Sin (90- 48°) + Cos 48° } { Sin(90-48°) - cos48*}
=> (Cos48° + Cos48° ) ( Cos48° - Cos48°)
=> (Cos²48°) - (Cos²48)
=> 0
HOPE IT WILL HELP YOU...... :-)
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