Question

How do you solve these ? 
a) Prove that for all values of x : (cos^4)x - (sin^4)x = (cos^2)x - (sin^2)x 
b) Given that 5(cos^2)x + 2(sin^2)x = 4, show that (tan^2)x = 1/2 
Urgent help needed. Thank you very much ! :)

Answer 1

a)
$a^{2}-b^2=(a-b)(a+b)$ formula used
$(sin^2x+cos^2x)=1$
$cos^4 x - sin^4 x=(sin^2x+cos^2x)(cos^2x-sin^2x)=(cos^2x-sin^2x)$
$(cos^2x-sin^2x)=cos2x$
b)
$5cos^2x + 2sin^2x = 4 \\ 5 \frac{cos^2x}{sin^2x}+2= \frac{4}{sin^2x} \\ 5cot^2x+2=4cosec^2x$
$cot^2x+2=4(cosec^2x-cot^2x)=4$
$(cosec^2x-cot^2x)=1$ formula
$cot^2x=2$
$cot x= \frac{1}{tan x}$ formula
$tan^2x= \frac{1}{2}$

Answer 2

a) Prove that for all values of x : (cos^4)x - (sin^4)x = (cos^2)x - (sin^2)x 
LHS+(cos^4)x - (sin^4)x=(cos²x)² - (sin²x)²
=(cos²x+sin²x)(cos²x - sin²x)
=(1)(cos²x - sin²x) =cos²x - sin²x =RHS
b) Given that 5(cos^2)x + 2(sin^2)x = 4, show that (tan^2)x = 1/2 
 5cos²x + 2sin²x =4
→3cos²x+2cos²x+ 2sin²x=4
→  3cos²x+2(cos²x+ sin²x)=4
→  3cos²x+2=4  or 3cos²x=2
i.e cos²x=2/3
and sin²x=1−cos²x =1−(2/3)=1−2/3 =1/3
Hence tan²x=sin²x/cos²x =(1/3)/(2/3)=1/2