Verified Answer :

1) tan^2(x)csc^2(x)-1/tan^2(x) 

= tan^2(x)
csc^2(x)/tan^2(x) - 1/tan^2(x) 

= csc^2(x)-cot^2(x) 

=(1+cot^2(x))-cot^2(x) 

= 1 

2) (cos x (sec x +1)/(sec x -1)(sec x +1)) + (cos x (sec x -1)/(sec x -1)(sec x +1)) 

= (1+ cos x)+ (1 - cos x)/(sec^2x-1) 
= 2/ tan^2(x) 

3) Find the missing side for triangle A and triangle B, 'a' lies in quadrant II it is positive, b is in quadrant III so it is negative. when you have found your sides plug it into the identity for tan(a-b) which is tan (a) - tan (b)/1+(tan a)(tan b), and simplify to find your answer. 

4) I think it is c. not sure though

caylus

Hello,

Then the answer i 1/(sin²θ)

Let's assume x=θ ( for the typo)

\(\dfrac{tan^2x+1}{tan^2x}\)

\(=1+\dfrac{cos^2x}{sin^2x}\)

\(=\dfrac{sin^2x+cos^2x}{sin^2x}\)

\(=\dfrac{1}{sin^2x}\)