How can we answer this question about identity?

1.   sin²(x) - 2  =  -sin(x)

This can't be an identity because it is not always true. For instance...

sin²(0) - 2  =  -sin(0)

0² - 2  =  -0

-2  =  0        This is false, so this equation is not an identity.

However, this equation does have solutions. We can find its solutions.

sin²(x) - 2  =  -sin(x)

                                             Add  sin(x)  to both sides of the equation.

sin²(x) + sin(x) - 2  =  0

                                             Factor the left side like this:  u² + u - 2  =  (u + 2)(u - 1)

(sin(x) + 2)(sin(x) - 1)  =  0

                                                             Set each factor equal to zero.

sin(x) + 2  =  0     or     sin(x) - 1  =  0

sin(x)  =  -2          or     sin(x)  =  1

sin  returns values between  -1  and  1  inclusively, so we can rule out the first option. That leaves...

sin(x)  =  1

x  =  arcsin(1)  =  90°

There will also be a solution every time  360°  is added to  90° , so all the real solutions are

x  =  90° + 360°n      where  n  is an integer

Does this answer your question? I might have misunderstood your question.

2.   (tan²x) / (1 + sec x)   =   sec x + 1      Let's test whether this is true when  x = 0 .

(tan²x) / (1 + 1/cosx)   =   1/cosx + 1

(tan²0) / (1 + 1/cos0)   =   1/cos0 + 1

(0²) / (1 + 1/1)   =   1/1 + 1

0  =  2       This is false, so this equation is not true for all values of  x  and so it is not an indentity.

If you want to find the solutions to the equation, we can try to find them.

(tan²x) / (1 + sec x)   =   sec x + 1       Multiply both sides by  (1 + sec x)

tan²x   =   (sec x + 1)(1 + sec x)

tan²x  =  sec x + sec²x + 1 + sec x

tan²x  =  sec²x + 2sec x + 1                Rewrite  tan  and  sec  in terms of  sin  and  cos.

sin²x / cos²x  =  1/cos²x + 2/cos x  +  1        Multiply through by  cos²x

sin²x  =  1 + 2cos x + cos²x                 Subtract  sin²x  from both sides.

0  =  1 + 2cos x + cos²x - sin²x           Substitute  1 - cos²x  in for  sin²x

0  =  1 + 2cos x + cos²x - (1 - cos²x)

0  =  1 + 2cos x + cos²x - 1 + cos²x

0  =  2cos x + 2cos²x         Divide through by  2 .

0  =  cos x + cos²x             Factor  cos x  out of both terms.

0  =  cos x(1 + cos x)         Set each factor equal to zero.

cos x  =  0      or      1 + cos x  =  0

However, neither of these can be true.

If cos x  =  0  then  sec x  is undefined and the original equation is undefined.

If  cos x  =  -1  then  1 + sec x  =  1 - 1  =  0  and the original equation is undefined.

So this equation has no solutions.