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| 1. |
If sinA =4/5,find the value of 16+16cot^2A |
| Answer» Let ABC be a right angle triangle,right angled at BTherefore sinA =BC/AC=4/5AC^2=BC^2+AB^225=16+AB^2AB^2=9AB=3CotA=3/416+16cot^2A=16+16(3/4)^2=16+16(9/16)=16+9=25 | |