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If `u=(1+cos theta)(1+cos 2theta)-sin theta.sin 2theta, v=sin theta(1+cos 2theta)+sin 2theta(1+cos theta)`, then `u^(2)+v^(2)=`A. `4(1+cos theta)(1+cos 2theta)`B. `4(1+sin theta)(1+sin 2theta)`C. `4(1-cos theta)(1-cos 2theta)`D. `4(1-sin theta)(1-sin 2theta)` |
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Answer» Correct Answer - A `u=1+cos theta + cos 2theta + cos (theta + 2theta)` `=(1+cos 3theta)+(cos theta + cos 2theta)` `=2 "cos"^(2)(3theta)/(2)+2 cos.(3theta)/(2)cos.(theta)/(2)` `=2 cos.(3theta)/(2)[cos.(3theta)/(2)+cos.(theta)/(2)]` Similarly `v=2 sin.(3theta)/(2)[cos.(3theta)/(2)+cos.(theta)/(2)]` `therefore u^(2)+v^(2)=4(cos.(3theta)/(2)+cos.(theta)/(2))^(2)` `=4(2cos theta. cos.(theta)/(2))^(2)=16 cos^(2)theta."cos"^(2)(theta)/(2)` `= 4(1+cos theta)(1+cos 2 theta)` |
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