Question

помогите пожалуйста 7 и 8​

Answer 1

Ответ:

$\displaystyle 7)\ \ (tga+tg\beta )(tga-tg\beta )-\Big(\frac{1}{cosa}+\frac{1}{cos\beta }\Big)\Big(\frac{1}{cosa}-\frac{1}{cos\beta }\Big)=\\\\\\=tg^2a-tg^2\beta -\Big(\frac{1}{cos^2a}-\frac{1}{cos^2\beta }\Big)=\frac{sin^2a}{cos^2a} -\frac{sin^2\beta }{cos^2\beta }-\frac{1}{cos^2a}+\frac{1}{cos^2\beta }=\\\\\\=\frac{sin^2a-1}{cos^2a}+\frac{1-sin^2\beta }{cos^2\beta }=\frac{-cos^2a}{cos^2a}+\frac{cos^2\beta }{cos^2\beta }=-1+1=0$

$8)\ \ cos^6\Big(a-\dfrac{\pi}{2}\Big)+sin^6\Big(a-\dfrac{3\pi}{2}\Big)-\dfrac{3}{4}\cdot \Big(sin^2\Big(a+\dfrac{\pi}{2}\Big)-cos^2\Big(a+\dfrac{3\pi}{2}\Big)\Big)^2=\\\\\\=sin^6a+cos^6a-\dfrac{3}{4}\cdot \Big(cos^2a-sin^2a\Big)^2=\\\\\\=\Big(\underbrace{sin^2a+cos^2a}_{1}\Big)\Big(sin^4a-sin^2a\cdot cos^2a+cos^4a\Big)-\\\\-\dfrac{3}{4}\cdot \Big(cos^4a-2\, cos^2a\cdot sin^2a+sin^4a\Big)=\\\\\\=sin^4a-sin^2a\cdot cos^2a+cos^4a-\dfrac{3}{4}\, cos^4a+\dfrac{3}{2}\, sin^2a\cdot cos^2a-\dfrac{3}{4}\, sin^4a=$

$=\dfrac{1}{4}\, sin^4a+\dfrac{1}{2}\, sin^2a\cdot cos^2a+\dfrac{1}{4}\, cos^4a=\\\\\\=\dfrac{1}{4}\cdot \Big(sin^4a+2\, sin^2a\cdot cos^2a+cos^4a\Big)=\dfrac{1}{4}\cdot \Big(sin^2a+cos^2a\Big)^2=\dfrac{1}{4}$

$\dfrac{1}{4}=\dfrac{1}{4}$