Question

с с заданием по алгебре
3 задание,докажите тождества

Answer 1

Объяснение:

3.

$a)\ \frac{cos^4\alpha -sin^4\alpha }{cos^2\alpha } +2tg^2\alpha =\frac{(cos^2\alpha -sin^2\alpha )*(cos^2\alpha +sin^2\alpha )}{cos^2\alpha } +2tg2\alpha =\\=\frac{(cos^2\alpha -sin^2\alpha)*1 }{cos^2\alpha } +2tg^2\alpha=\frac{cos^2\alpha -sin^2\alpha }{cos^2\alpha } +2tg^2\alpha= 1-tg^2\alpha +2tg^2\alpha =\\=1+tg^2\alpha =1+\frac{sin^2\alpha }{cos^2\alpha } =\frac{cos^2\alpha+sin^2\alpha }{cos^2\alpha } =\frac{1}{cos^2\alpha } .$

$b)\ \frac{tg\alpha }{tg\alpha +ctg\alpha } =\frac{tg\alpha }{\frac{sin\alpha }{cos\alpha }+\frac{cos\alpha }{sin\alpha } } =\frac{\frac{sin\alpha }{cos\alpha } }{\frac{sin^2\alpha +cos^2\alpha }{sin\alpha *cos\alpha } } =\frac{sin\alpha }{\frac{1}{sin\alpha } } =sin^2\alpha .$

$a)\ \frac{sin^4\alpha -cos^4\alpha }{sin^2\alpha } +2ctg^2\alpha =\frac{(sin^2\alpha -cos^2\alpha )*(sin^2\alpha +cos^2\alpha )}{sin^2\alpha } +2ctg^2\alpha =\\=\frac{(sin^2\alpha -cos^2\alpha)*1 }{sin^2\alpha } +2ctg^2\alpha =\frac{sin^2\alpha -cos^2\alpha }{sin^2\alpha } +2ctg^2\alpha =1-ctg^2\alpha +2ctg^2\alpha =\\=1+ctg^2\alpha =1+\frac{cos^2\alpha }{sin^2\alpha } =\frac{sin^2\alpha +cos^2\alpha }{sin^2\alpha } =\frac{1}{sin^2\alpha } .$

$b)\ \frac{ctg\alpha }{tg\alpha +ctg\alpha } =\frac{ctg\alpha }{\frac{sin\alpha }{cos\alpha} +\frac{cos\alpha }{sin\alpha } } } =\frac{\frac{cos\alpha }{sin\alpha } }{\frac{sin^2\alpha +cos^2\alpha }{sin\alpha *cos\alpha } } =\frac{cos\alpha }{\frac{1}{cos\alpha } } =cos^2\alpha .$

Answer 2

Использовали формулы:   $sin^2a+cos^2a=1\ \ ,\ \ tga=\dfrac{sina}{cosa}\ \ ,\ \ ctga=\dfrac{cosa}{sina}\ \ ,\ \ a^2-b^2=(a-b)(a+b)$  

$\displaystyle 1)\ \ \frac{cos^4a-sin^4a}{cos^2a}+2tg^2a=\frac{1}{cos^2a}frac{cos^4a-sin^4a}{cos^2a}+2tg^2a=\frac{(cos^2a-sin^2a)(\overbrace{cos^2a+sin^2a}^{1})}{cos^2a}+2tg^2a==\frac{cos^2a-sin^2a}{cos^2a}+2tg^2a=\frac{cos^2a}{cos^2a}-\frac{sin^2a}{cos^2a}+2tg^2a=1-tg^2a+2tg^2a==1+tg^2a=\frac{1}{cos^2a}frac{1}{cos^2a}=\frac{1}{cos^2a}$

$\displaystyle 2)\ \ \frac{tga}{tga+ctga}=sin^2afrac{tga}{tga+ctga}=\frac{\dfrac{sina}{cosa}}{\dfrac{sina}{cosa}+\dfrac{cosa}{sina}}=\frac{sina\cdot cosa\cdot sina}{cosa\, (\underbrace{sin^2a+cos^2a}_{1})}=sin^2asin^2a=sin^2a$

$\displaystyle 3)\ \ \frac{sin^4a-cos^4a}{sin^2a}+2ctg^2a=\frac{1}{sin^2a}frac{sin^4a-cos^4a}{sin^2a}+2ctg^2a=\frac{(sin^2a-cos^2a)(\overbrace{sin^2a+cos^2a}^{1})}{sin^2a}+2ctg^2a==\frac{sin^2a-cos^2a}{sin^2a}+2ctg^2a=\frac{sin^2a}{sin^2a}-\frac{cos^2a}{sin^2a}+2ctg^2a=1-ctg^2a+2ctg^2a==1+ctg^2a=\frac{1}{sin^2a}frac{1}{sin^2a}=\frac{1}{sin^2a}$

$\displaystyle 4)\ \ \frac{ctga}{tga+ctga}=cos^2afrac{ctga}{tga+ctga}=\frac{\dfrac{cosa}{sina}}{\dfrac{sina}{cosa}+\dfrac{cosa}{sina}}=\frac{cosa\cdot cosa\cdot sina}{sina\, (\underbrace{sin^2a+cos^2a}_{1})}=cos^2acos^2a=cos^2a$