![sinalpha+sinbeta-sin(alpha+beta))/(sinalpha+sinbeta+sin(alpha+beta))=tan( alpha/2)tan(beta/2)` - YouTube sinalpha+sinbeta-sin(alpha+beta))/(sinalpha+sinbeta+sin(alpha+beta))=tan( alpha/2)tan(beta/2)` - YouTube](https://i.ytimg.com/vi/aqYyx-vO3QU/maxresdefault.jpg)
sinalpha+sinbeta-sin(alpha+beta))/(sinalpha+sinbeta+sin(alpha+beta))=tan( alpha/2)tan(beta/2)` - YouTube
![If a = [(Cos Alpha, Sin Alpha), (-sin Alpha, Cos Alpha)] Then Verify That A'A = I - Mathematics | Shaalaa.com If a = [(Cos Alpha, Sin Alpha), (-sin Alpha, Cos Alpha)] Then Verify That A'A = I - Mathematics | Shaalaa.com](https://www.shaalaa.com/images/_4:c824d6fd2090415e8026edbb0c866a20.png)
If a = [(Cos Alpha, Sin Alpha), (-sin Alpha, Cos Alpha)] Then Verify That A'A = I - Mathematics | Shaalaa.com
![trigonometry - Relation between $\sin(\cos(\alpha))$ and $\cos(\sin(\alpha))$ - Mathematics Stack Exchange trigonometry - Relation between $\sin(\cos(\alpha))$ and $\cos(\sin(\alpha))$ - Mathematics Stack Exchange](https://i.stack.imgur.com/TONvu.png)
trigonometry - Relation between $\sin(\cos(\alpha))$ and $\cos(\sin(\alpha))$ - Mathematics Stack Exchange
![The expression `(sin(alpha+theta)-\""sin\""(alpha-theta))/(cos(beta-theta)-cos(beta+theta))`is -ind - YouTube The expression `(sin(alpha+theta)-\""sin\""(alpha-theta))/(cos(beta-theta)-cos(beta+theta))`is -ind - YouTube](https://i.ytimg.com/vi/SIKrOxVh7GI/maxresdefault.jpg)
The expression `(sin(alpha+theta)-\""sin\""(alpha-theta))/(cos(beta-theta)-cos(beta+theta))`is -ind - YouTube
![If sin,alpha +sin,beta = a and cos,alpha +cos,beta = b, show that cos (alpha +beta ) = dfrac{b^2-a^2}{b^2+a^2} If sin,alpha +sin,beta = a and cos,alpha +cos,beta = b, show that cos (alpha +beta ) = dfrac{b^2-a^2}{b^2+a^2}](https://haygot.s3.amazonaws.com/questions/1410367_1664496_ans_0b1fb2b3811b4b359d0ad6fd0e8cf14b.jpg)
If sin,alpha +sin,beta = a and cos,alpha +cos,beta = b, show that cos (alpha +beta ) = dfrac{b^2-a^2}{b^2+a^2}
![Prove that dfrac{sin alpha +sin beta}{sinalpha -sinbeta}=dfrac{tan dfrac{ alpha +beta}{2}}{tandfrac{alpha -beta}{2}}. Prove that dfrac{sin alpha +sin beta}{sinalpha -sinbeta}=dfrac{tan dfrac{ alpha +beta}{2}}{tandfrac{alpha -beta}{2}}.](https://haygot.s3.amazonaws.com/questions/1556027_1729848_ans_3fa22f573a98433aa402386217820239.jpg)
Prove that dfrac{sin alpha +sin beta}{sinalpha -sinbeta}=dfrac{tan dfrac{ alpha +beta}{2}}{tandfrac{alpha -beta}{2}}.
![If a = [(Sin Alpha, Cos Alpha),(-cos Alpha, Sin Alpha)] Then Verify That A'A = I - Mathematics | Shaalaa.com If a = [(Sin Alpha, Cos Alpha),(-cos Alpha, Sin Alpha)] Then Verify That A'A = I - Mathematics | Shaalaa.com](https://www.shaalaa.com/images/_4:224fcb85876c471889528b2b269be222.png)
If a = [(Sin Alpha, Cos Alpha),(-cos Alpha, Sin Alpha)] Then Verify That A'A = I - Mathematics | Shaalaa.com
![If sin \alpha= 3/5, \sin \beta= 5/13, \alpha is in quadrant I and \beta is in quadrant II, find \sin(\alpha +\beta). | Homework.Study.com If sin \alpha= 3/5, \sin \beta= 5/13, \alpha is in quadrant I and \beta is in quadrant II, find \sin(\alpha +\beta). | Homework.Study.com](https://homework.study.com/cimages/multimages/16/hypotenuse2272442476950271681.png)