Question

Which of the following is an identity? A. sin2x sec2x + 1 = tan2x csc2x B. sin2x - cos2x = 1 C. (cscx + cotx)2 = 1 D. csc2x + cot2x = 1
Edit: no one answered but I figured out that the right answer is A

Answer 1

Answer: A is the correct answer.

Step-by-step explanation:

Just did the question, followed OP's advice since no one answered. Good luck.

Answer 2

There are three 'Pythagorean' identities that we can look at and they are

sin²(x) + cos²(x) = 1
tan²(x) + 1 = sec²(x) 
1 + cot²(x) = csc²(x)

We can start by checking each option to see which one would give us any of the 'Pythagorean' identities as its simplest form

Option A:

sin²(x) sec²(x) + 1 = tan²(x) csc²(x)

Rewriting sec²(x) as 1/cos²(x)
Rewriting tan²(x) as sin²(x)/cos²(x)
Rewriting csc²(x) as 1/sin²(x)

We have

$sin^{2}(x)[ \frac{1}{ cos^{2}(x) }]+1=[ \frac{ sin^{2}( x)}{ cos^{2} (x)}][ \frac{1}{ sin^{2}(x) } ]$
$[\frac{ sin^{2}(x) }{ cos^{2}(x) } ]+1= \frac{1}{ cos^{2}(x) }$
$tan^{2}(x)+1= sec^{2}(x)$

Option B:

sin²(x) - cos²(x) = 1

This expression is already in the simplest form, cannot be simplified further

Option C:

[ csc(x) + cot(x) ]² = 1

Rewriting csc(x) as 1/sin(x)
Rewriting cot(x) as cos(x)/sin(x)

We have

$[ \frac{1}{sin(x)}+ \frac{cos(x)}{sin(x)}] ^{2} =1$
$\frac{1}{sin^2(x)}+2( \frac{1}{sin(x)})( \frac{cos(x)}{sin(x)})+ \frac{cos^2(x)}{sin^2(x)}=1$$csc^2(x)+2csc^2(x)cos(x)+cot^2(x)=1$

Option D:

csc²(x) + cot²(x) = 1

Rewriting csc²(x) as 1/sin²(x) and cot²(x) as cos²(x)/sin²(x)

$\frac{1}{sin^2(x)}+ \frac{cos^2(x)}{sin^2(x)}=1$
$\frac{1+cos^2(x)}{sin^2(x)} =1$
$1+cos^2(x)=sin^2(x)$
$1=sin^2(x)-cos^2(x)$

from our working out we can see that option A simplified into one of 'Pythagorean' identities, hence the correct answer