Question

$\text{If}\text{tan}\text{(A + B) =}\sqrt{3};\tan \text{(A - B) =}\frac{1}{\sqrt{3}}$ Solve for A & B.
Also Simplify :
$\frac{\left(1+\sin \theta \right)\left(1-\sin \theta \right)}{\left(1+\cos \theta \right)\left(1-\cos \theta \right)}$

Answer 1

Solution :- given

$tan(A+B)=\sqrt{3}$ $tan(A-B)=\frac{1}{\sqrt{3}}$

$A+B=\frac{\pi }{3}$ $A-B=\frac{\pi }{6}$

finding A an B By elimination

$\Rightarrow A+B=\frac{\pi }{3}$

$A-B=\frac{\pi }{6}$

___________________

$2A=\frac{\pi }{3}+\frac{\pi }{6}$

$A=\frac{2\pi +\pi }{6\times 2}=\frac{3\pi }{6\times 2}=\frac{\pi }{4}$

$B=\frac{\pi }{3}-\frac{\pi }{4}=\frac{4\pi -3\pi }{12}=\frac{\pi }{12}$

Now given

$\Rightarrow \frac{(1+sin\theta )(1-sin\theta )}{(1+cos\theta )(1-cos\theta )}$

$=\frac{1-si{n}^2\theta }{1-co{s}^2\theta }$

$=\frac{co{s}^2\theta }{si{n}^2\theta }$

$=co{t}^2\theta$