Question

please answer the question

Answer 1

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Answer 2

Given, cos theta + sin theta = rt2 sin theta
On squaring L.H.S & R.H.S
(cos theta + sin theta)^2 = (rt 2 sin theta)^2
cos^2 theta + sin^2 theta + 2 sin theta cos theta = 2 sin^2 theta
cos^2 theta - sin^2 theta + 2 sin theta cos theta = 0
cos^2 theta - sin^2 theta = - 2 sin theta cos theta..... (i)

cos^2 theta - sin^2 theta = - 2 sin theta cos theta
cos^2 theta - sin^2 theta + 2 sin theta cos theta -2cos^2theta = -2cos^2theta
-cos^2 theta - sin^2 theta + 2 sin theta cos theta = -2cos^2theta
-(cos^2 theta + sin^2 theta - 2 sin theta cos theta) = -2cos^2theta
(cos theta - sin theta )^2 = 2cos^2theta
(cos theta - sin theta ) = rt 2 cos theta....

{Alternatively, after steps up-to equation (i) part, you can follow this too:
Now: (cos theta - sin theta)^2
(cos theta - sin theta)^2 =cos^2 theta + sin^2 theta - 2 sin theta cos theta
Sub (i) in above for - 2 sin theta cos theta
(cos theta - sin theta)^2 =cos^2 theta + sin^2 theta + cos^2 theta - sin^2 theta
(cos theta - sin theta)^2 = 2 cos^2 theta
Take square root on both sides:
(cos theta - sin theta) = rt2 cos theta }

Please check the question part again. Either one of them has to be rt 2 cos theta on the R.H.S part, i guess.