If you could go step by step it would be helpful!
cotx+tanx = secxcscx
sin^4x - cos^4x = 2sin^2x - 1
cot^2x-1/1+cot^2x = 1-2sin^2x
Thanks!
$\displaystyle \frac{cot^2(x)-1}{cot^2(x)+1}$
$\displaystyle \frac{cot^2(x)-1+1-1}{cot^2(x)+1}$
$\displaystyle \frac{cot^2(x)+1-2}{cot^2(x)+1}$
$\displaystyle \frac{cot^2(x)+1}{cot^2(x)+1}+\frac{-2}{cot^2(x)+1}$
$\displaystyle 1+\frac{-2}{cot^2(x)+1}$
$\displaystyle 1+\frac{-2}{csc^2(x)}$
$\displaystyle 1-2sin^2(x)$
Q1
Write LHS in terms of sin and cos
Get a common denominator and add
You should recognise and simplify the top
Split up the product
You should now have the RHS
TRY IT!!
Q2 Hint: a^4 - b^4 =(a^2 + b^2)(a^2 - b^2)
The first bracket will be =1
Look at the RHS (there is no cos in sight!)
So, replace (cos x)^2 with 1 - (sin x)^2
Too easy!
Q3 Use the identity 1 + (cot x)^2 = (cosec x)^2 on the bottom
Dividing by (cosec x)^2 is the same as multiplying by (sin x)^2
So now you should have (sin x)^2 x ((cot x)^2 -1)
Write cot^2 in terms of sin and cos
Expand out bracket
Do the same thing you did at the end of Q2 to get rid of cos^2
Done
$\displaystyle cot(x)+tan(x)=sec(x)csc(x)$
$\displaystyle \frac{cos(x)}{sin(x)}+\frac{sin(x)}{cos(x)}$
You choose the common denominator of $\displaystyle cos(x)sin(x)$ and multiply through. You get:
$\displaystyle \frac{cos^2(x)+sin^2(x)}{sin(x)cos(x)}$
Use the identity $\displaystyle sin^2(x)+cos^2(x)=1$
$\displaystyle \frac{1}{sin(x)cos(x)}$
$\displaystyle \frac{1}{sin(x)}*\frac{1}{cos(x)}$
$\displaystyle csc(x)sec(x)$
Edit: If you still need me to show you the second one after Debsta explained it, post the request. Debsta did a good job of explaining it.