If (cos theta-sin theta)=root of 2 sin theta,prove that cos theta +sin theta = root of 2 cos theta..............Please guys somebody answer this very urgent
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Trigonometry,
We have,
using e insted of theta ok,
cose - sine = √2sine
Have to prove that,
cose + sine = √2cose.
Now,
cose - sine = √2sine
= cose = √2sine + sine
= cose = (√2 + 1)sine
= cose/(√2 + 1) = sine
= {cose×(√2 - 1)}/{(√2)²-1} = sine
= √2cose - cose = sine
= √2cose = sine + cose [ proved ]
That's it
Hope it helped (●´ϖ`●)
We have,
using e insted of theta ok,
cose - sine = √2sine
Have to prove that,
cose + sine = √2cose.
Now,
cose - sine = √2sine
= cose = √2sine + sine
= cose = (√2 + 1)sine
= cose/(√2 + 1) = sine
= {cose×(√2 - 1)}/{(√2)²-1} = sine
= √2cose - cose = sine
= √2cose = sine + cose [ proved ]
That's it
Hope it helped (●´ϖ`●)
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