If cosα+cosβ=m and sinα+sinβ=n, then sin(α+β)=

# If and $\mathrm{sin}\alpha +\mathrm{sin}\beta =n,$ then $\mathrm{sin}\left(\alpha +\beta \right)=$

1. A

$\frac{{m}^{2}-{n}^{2}}{{m}^{2}+{n}^{2}}$

2. B

$\frac{{n}^{2}-{m}^{2}}{{m}^{2}+{n}^{2}}$

3. C

$\frac{2mn}{{m}^{2}+{n}^{2}}$

4. D

$\frac{2mn}{{m}^{2}-{n}^{2}}$

Register to Get Free Mock Test and Study Material

+91

Verify OTP Code (required)

### Solution:

$m=\mathrm{cos}\alpha +\mathrm{cos}\beta =2\mathrm{cos}\left(\frac{\alpha +\beta }{2}\right)\mathrm{cos}\left(\frac{\alpha -\beta }{2}\right)$

$n=\mathrm{sin}\alpha +\mathrm{sin}\beta =2\mathrm{sin}\left(\frac{\alpha +\beta }{2}\right)\mathrm{cos}\left(\frac{\alpha -\beta }{2}\right)$

Now,

$\mathrm{sin}\left(\alpha +\beta \right)=\frac{2\mathrm{tan}\left(\frac{\alpha +\beta }{2}\right)}{1+{\mathrm{tan}}^{2}\left(\frac{\alpha +\beta }{2}\right)}=\frac{\frac{2n}{m}}{1+\left(\frac{{n}^{2}}{{m}^{2}}\right)}=\frac{2mn}{{m}^{2}+{n}^{2}}$

Register to Get Free Mock Test and Study Material

+91

Verify OTP Code (required)