Question

15. If m sin(0 + a) = n sin(0 + B) then show that cot 0 =
mcosa - ncos B
nsin ß- msin a
x sino
y sin e
sin Ꮎ
1 - x cos o
then show that
1- y cose
y sin o
tan o =
16. If tan 0 =
1.​

Answer 1

Answer:

I don't know it is hard...

Answer 2

Q: if  $m sin (\theta + \alpha)=n sin(\theta + \beta)$  then show that  $cos \theta=\frac{mcos\alpha - ncos\beta}{nsin\beta - msin \alpha}$

Answer:

Given

$m sin (\theta + \alpha)=n sin(\theta + \beta)$

Dividing both sides by $cos\theta$ :

$\frac{m sin(\theta + \alpha) }{cos \theta}= \frac{n sin(\theta + \beta )}{cos \theta}\\ \\\frac{m[sin \theta cos \alpha + cos \theta sin \alpha]}{cos \theta}=\frac{n [sin \theta cos \beta + cos \theta sin \beta]}{cos \theta}\\\\ m[\frac{sin \theta cos \alpha}{cos \theta} + \frac{cos \theta sin \alpha}{cos \theta} ]= n [\frac{sin \theta cos \beta}{cos \theta} + \frac{cos \theta sin \beta}{cos \theta} ]\\\\m[tan \theta cos \alpha + sin \alpha]=n[tan \theta cos\beta + sin\beta]$

$\\ mtan\theta cos \alpha + msin\alpha=ntan\thetacos\beta + nsin\beta\\\\mtan\theta cos\alpha-ntan\theta cos\beta=nsin\beta - msin\alpha\\\\tan\theta [mcos \alpha - ncos\beta]=nsin\beta - msin\alpha$

$\frac{1}{cot \theta}=\frac{nsin\beta - msin \alpha}{mcos\alpha - ncos\beta}\\ \\ cot \theta=\frac{mcos\alpha - ncos\beta}{nsin\beta - msin \alpha}$