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If `3 cot theta= 2` , show that `((4 sin theta - 3 cos theta ))/((2 sin theta + 6 cos theta))=(1)/(3)` |
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Answer» `((4sin theta-3 cos theta ))/((2 sin theta +6 cos theta ))=((4-3cot theta ))/((2+ 6 cot theta))" " ["dividing num, and denom , by " sin theta ]` `=((4-3xx(2)/(3)))/((2+6xx(2)/(3)))=(2)/(6)=(1)/(3).` |
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