Prove the identity
Tan²x - sec²x + 2 = cosec²x - cot²x
To prove the identity:
Tan²x - sec²x + 2 = cosec²x - cot²x
We will start by expanding and simplifying each side of the equation separately:
On the left side, we have:
Tan²x - sec²x + 2
= (sin²x / cos²x) - (1 / cos²x) + 2 [Using the definitions of tangent (Tanx) and secant (secx)]
= (sin²x - 1 + 2cos²x) / cos²x [Finding the common denominator and combining the terms]
= (sin²x + 2cos²x - 1) / cos²x [Rearranging the terms]
On the right side, we have:
Cosec²x - cot²x
= (1 / sin²x) - (cos²x / sin²x) [Using the definitions of cosecant (cosecx) and cotangent (cotx)]
= (1 - cos²x) / sin²x [Finding the common denominator and combining the terms]
= (sin²x) / sin²x [Using the Pythagorean identity: sin²x + cos²x = 1]
= 1 [Simplifying]
Therefore, we have:
(sin²x + 2cos²x - 1) / cos²x = 1
Now, let's simplify the expression further:
(sin²x + 2cos²x - 1) = cos²x
Using the Pythagorean identity: sin²x = 1 - cos²x, we can substitute it into the equation:
(1 - cos²x + 2cos²x - 1) = cos²x
cos²x + cos²x - 1 = cos²x
2cos²x - 1 = cos²x
2cos²x - cos²x = 1
cos²x = 1
Since cosine (cosx) is always between -1 and 1, the only possible value for cos²x to equal 1 is when cosx = ±1. Therefore, the original equation holds true for all values of x where cosx = ±1.
Thus, the identity is proved for all values of x where cosx = ±1.
To prove the identity:
Tan²x - sec²x + 2 = cosec²x - cot²x
Let's start with the left-hand side (LHS):
LHS = Tan²x - sec²x + 2
Using the Pythagorean Identity:
1 + Tan²x = sec²x
LHS = (1 + Tan²x) - sec²x + 2
Combining like terms, we get:
LHS = 1 + Tan²x - sec²x + 2
Now, let's focus on the right-hand side (RHS):
RHS = cosec²x - cot²x
Using the reciprocal identities:
cosec²x = 1 + cot²x
RHS = (1 + cot²x) - cot²x
Again, combining like terms:
RHS = 1 + cot²x - cot²x
Simplifying further, we get:
RHS = 1
Since LHS = 1 + Tan²x - sec²x + 2 and RHS = 1, we have successfully shown that:
Tan²x - sec²x + 2 = cosec²x - cot²x