Answered By Safian Q

1a. Recall that  $sec^2x=\frac{1}{cos^2x}$sec²x=1cos²x  . We can't let  $cos^2x=0$cos²x=0 because  $\frac{1}{0}=undefined$10 =undeƒ ined . For what values does    $cos^2x=0$cos²x=0? Think about where just $cosx=0$cosx=0 , which is at  $\frac{1\pi}{2},\frac{3\pi}{2},\frac{5\pi}{2}...$1π2 ,3π2 ,5π2 ... You can write this as  $\frac{n\pi}{2}$nπ2  where the value of  $n$n is an odd number. Those are the non permissible values of the identiy. 

1b. I assume here we just plug in a number for $x$x to check if both sides equal. Let's plug $0$0 for $x$x on the lefthand side first:

 $tan^20\left(cos^20+1\right)+cos^20=0\left(1+1\right)+1^2=1$tan²0(cos²0+1)+cos²0=0(1+1)+1²=1 

Now let's do the righthand side:

 $sec^20=\frac{1}{cos^20}=\frac{1}{1^2}=1$sec²0=1cos²0 =11² =1 

so you can see they're equal.

1c. Graph (they're overlapping eachother so which shows they're equal):

1d. Proof:

 $tan^2x\left(cos^2x+1\right)+cos^2x=sec^2x$tan²x(cos²x+1)+cos²x=sec²x 

 $\frac{sin^2x}{cos^2x}\left(cos^2x+1\right)+cos^2x=\frac{1}{cos^2x}$sin²xcos²x (cos²x+1)+cos²x=1cos²x  , using definition of tan and secant. 

 $sin^2x+\frac{sin^2x}{cos^2x}+cos^2x=\frac{1}{cos^2x}$sin²x+sin²xcos²x +cos²x=1cos²x   , multiplying the brackets

$1+\frac{sin^2x}{cos^2x}=\frac{1}{cos^2x}$1+sin²xcos²x =1cos²x  , using  $sin^2x+cos^2x=1$sin²‍x+cos²x=1 

 $\frac{\left(cos^2x+sin^2x\right)}{cos^2x}=\frac{1}{cos^2x}$(cos²x+sin²x)cos²x =1cos²x  , rearranging terms/denominator

 $\frac{1}{cos^2x}=\frac{1}{cos^2x}$1cos²x =1cos²x , using  $sin^2x+cos^2x=1$sin²x+cos²x=1 again