Question

Помогите с вычислением, пожалуйста

Answer 1

Ответ:

$1)\ \ \ tga=\dfrac{2}{3}\ \ ,\ \ tg\beta =\dfrac{1}{3}\\\\\\tg(a+\beta )=\dfrac{tga+tg\beta }{1-tga\cdot tg\beta }=\dfrac{\frac{2}{3}+\frac{1}{3}}{1-\frac{2}{3}\cdot \frac{1}{3}}=\dfrac{1}{1-\frac{2}{9}}=\dfrac{9}{7}=1\dfrac{2}{9}$

$2)\ \ tga=\dfrac{3}{2}\ \ ,\ \ tg\beta =\dfrac{3}{2}\ \ \ \to \ \ \ ctga=ctg\beta =\dfrac{2}{3}\\\\\\ctg(a-\beta)=\dfrac{1+ctga\cdot ctg\beta }{ctga-ctg\beta }=\dfrac{1+\frac{2}{3}\cdot \frac{2}{3}}{\underbrace {\frac{2}{3}-\frac{2}{3}}_{0}}\ \ \ \ ne\ syshestvyet$

$3)\ \ sina=\dfrac{12}{13}\ \ ,\ \ \dfrac{\pi}{2}<\beta <\pi \\\\1+tg^2a=\dfrac{1}{cos^2a}\ \ ,\\\\tg^2a=\dfrac{1}{cos^2a}-1=\dfrac{1}{1-sin^2a}-1=\dfrac{1}{1-\frac{144}{169}}-1=\dfrac{169}{25}-1=\dfrac{144}{25}\ \ ,\\\\\dfrac{\pi}{2}<a<\pi \ \ \to \ \ tga<0\ \ ,\ \ \ tga=-\dfrac{12}{5}\\\\tg(\dfrac{\pi}{4}+\alpha )=\dfrac{tg\frac{\pi}{4}+tg\alpha }{1-tg\frac{\pi}{4}\cdot tga}=\dfrac{1-\frac{12}{5}}{1+1\cdot \frac{12}{5}}=\dfrac{-7}{17}$