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`        If cos (α+β) = 4/5 and sin (α-β) = 5/13Then tan 2α =?`
one year ago Deepak Kumar Shringi
4397 Points
``` ```
one year ago
```							Cos(α+β)=4/5, Tan(α+β)=3/4Sin(α-β)=5/13, Tan(α-β)=5/12Tan(2α)= tan[(α+β)+(α-β)]              =[tan(α+β)+tan(α-β)]/1-tan(α+β)tan(α-β)              =(3/4 + 5/12)/1-(3/4)(5/12)              =[(36+20)/48]/[(48-15)/48]              =56/33
```
one year ago Deepak Kumar Shringi
4397 Points
```							by mistake i have taken tan alpha+beta as ¾ but the method is same
```
one year ago
```							 Cos(α+β)=4/5, Tan(α+β)=3/4Sin(α-β)=5/13, Tan(α-β)=5/12Tan(2α)= tan[(α+β)+(α-β)]              =[tan(α+β)+tan(α-β)]/1-tan(α+β)tan(α-β)              =(3/4 + 5/12)/1-(3/4)(5/12)              =[(36+20)/48]/[(48-15)/48]              =56/33
```
one year ago
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