If cos A + sin A = √2 cos A then show that cos A -sin A =√2sin A
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Answered by
55
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Here is your answer
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Q.If cos A + sin A = √2 cos A then show that cos A -sin A =√2sin A
Solution :-
We have,
cos A + sin A =√2 cos A
=> ( cos A + sin A )² =2 cos² A
=> cos² A + sin² A + 2cos A sin A =2 cos²
=> cos² A - 2cos A sin A = sin² A
=> cos² A - 2cos A sin A + sin² A =2 sin ²A
=> ( cos A - sin A )² = 2 sin² A
=> cos A - sin A = √ 2 sin A
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Here is your answer
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Q.If cos A + sin A = √2 cos A then show that cos A -sin A =√2sin A
Solution :-
We have,
cos A + sin A =√2 cos A
=> ( cos A + sin A )² =2 cos² A
=> cos² A + sin² A + 2cos A sin A =2 cos²
=> cos² A - 2cos A sin A = sin² A
=> cos² A - 2cos A sin A + sin² A =2 sin ²A
=> ( cos A - sin A )² = 2 sin² A
=> cos A - sin A = √ 2 sin A
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Answered by
32
Heya !!!
Cos A + Sin A = ✓2 Cos A
=> ( Cos A + Sin A )² = 2 Cos² A [ On squaring both sides]
=> Cos² A + Sin² A + 2 Cos A Sin A = 2 Cos² A
=> Cos² A - 2 Cos A Sin A = Sin ²
=> Cos²A - 2 Cos A Sin A + Sin²A = Sin² A + Sin²A [ Adding Sin²A both sides]
=> ( Cos A - Sin A)² = 2 Sin² A
=> (Cos A - Sin A ) = ✓2 Sin A.
Hence,
(Cos A - Sin A) = ✓2 Sin A.....PROVED.......
HOPE IT WILL HELP YOU...... :-)
Cos A + Sin A = ✓2 Cos A
=> ( Cos A + Sin A )² = 2 Cos² A [ On squaring both sides]
=> Cos² A + Sin² A + 2 Cos A Sin A = 2 Cos² A
=> Cos² A - 2 Cos A Sin A = Sin ²
=> Cos²A - 2 Cos A Sin A + Sin²A = Sin² A + Sin²A [ Adding Sin²A both sides]
=> ( Cos A - Sin A)² = 2 Sin² A
=> (Cos A - Sin A ) = ✓2 Sin A.
Hence,
(Cos A - Sin A) = ✓2 Sin A.....PROVED.......
HOPE IT WILL HELP YOU...... :-)
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