KeerthanaPosted on The value of cosα+cosβsinα+sinβ\frac{\cos \alpha+\cos \beta}{\sin \alpha+\sin \beta}sinα+sinβcosα+cosβ will be atan(α+β2)\tan \left(\frac{\alpha+\beta}{2}\right)tan(2α+β) bcot(α+β2)\cot \left(\frac{\alpha+\beta}{2}\right)cot(2α+β) ctan(α−β2)\tan \left(\frac{\alpha-\beta}{2}\right)tan(2α−β) dcot(α−β2)\cot \left(\frac{\alpha-\beta}{2}\right)cot(2α−β) Answer : Option BExplanation : cosα+cosβsinα+sinβ=2⋅cos(α+β2)⋅cos(α−β2)2sin(α+β2)⋅cos(α−β2)\frac{\cos \alpha+\cos \beta}{\sin \alpha+\sin \beta}=\frac{2 \cdot \cos \left(\frac{\alpha+\beta}{2}\right) \cdot \cos \left(\frac{\alpha-\beta}{2}\right)}{2 \sin \left(\frac{\alpha+\beta}{2}\right) \cdot \cos \left(\frac{\alpha-\beta}{2}\right)}sinα+sinβcosα+cosβ=2sin(2α+β)⋅cos(2α−β)2⋅cos(2α+β)⋅cos(2α−β) =cos(α+β2)sin(α+β2)=cot(α+β2)=\frac{\cos \left(\frac{\alpha+\beta}{2}\right)}{\sin \left(\frac{\alpha+\beta}{2}\right)}=\cot \left(\frac{\alpha+\beta}{2}\right)=sin(2α+β)cos(2α+β)=cot(2α+β) Rate This:NaN / 5 - 1 votesAdd comment
