Chứng minh rằng : a) $\frac{\sqrt{1+cosa}+\sqrt{1-cosa}}{{1+cosa}-\sqrt{1-cosa}}$`=cot(a/2+pi/4)` `pi<a<2pi` b) `{cos4a tan 2a-sin4a}/{cos4a tan 2a+sin4a}` c)` 1+2cos7a={sin10,5a}/{sin3,5a}` d) {tan3a}/{tana}={3-tan^2a}/{1-3tan^2a}

Đáp án+Giải thích các bước giải:

`a)`

Vì `\sqrt{1+cos a}=-\sqrt{2}cos` `{a}/{2}(`do `pi/2<a/2<pi)`

`\sqrt{1-cos a}=``\sqrt{2}sin``a/2` cho nên

$\dfrac{\sqrt{1+cos a}+\sqrt{1-cosa}}{\sqrt{1+cosa}-\sqrt{1-cosa}}$`=` $\dfrac{-\sqrt{2}cos\dfrac{a}{2}+\sqrt{2}sin\dfrac{a}{2} }{-\sqrt{2}cos\dfrac{a}{2}-\sqrt{2}sin\dfrac{a}{2}}$

`=`$\dfrac{cos\dfrac{a}{2}-sin\dfrac{a}{2}}{cos\dfrac{a}{2}+sin\dfrac{a}{2}}$`=`$\dfrac{1-tan\dfrac{a}{2}}{1+tan\dfrac{a}{2}}$

`=tan(pi/4-a/2)=cot(a/2+pi/4)`

`b)`

`{cos4atan2a-sin4a}/{cos4acot2a+sin4a}={cos4asin2a-sin4acos2a}/{cos4acos2a+sin4asin2a}.tan2a`

`={-sin2a}/{cos2a} tan 2a=-tan^2 2a`

`c)`

`{sin10,5a}/{sin3,5a}={sin(7a+3,5a)}/{sin3,5a}`

`={sin7acos3,5a+cos7asin3,5a}/{sin3,5a}`

`={sin3,5a(2cos^2 3,5a+cos7a)}/{sin3,5a}`

`=(2cos^2 3,5a-1)+1cos7a=2cos7a+1`

`d)`

`tan(a+2a)/tana={tana+tan2a}/{tana(1-tanatan2a)}`

`={tana+{2tana}/{1-tan^2a}}/{tana(1-{2tan^2a}/{1-tan^2a})}={3-tan^2a}/{1-3tan^2a}`