To prove the given equation [tex]\(\operatorname{cosec}^2 A\left(1 + \cos^2 A\right) = t^2 \cot^2 A\)[/tex], we will rely on fundamental trigonometric identities and algebraic manipulation.
### Step-by-Step Solution
#### Step 1: Express all trigonometric functions in terms of sine and cosine
Recall the trigonometric identities:
1. [tex]\(\operatorname{cosec} A = \frac{1}{\sin A}\)[/tex]
2. [tex]\(\cot A = \frac{\cos A}{\sin A}\)[/tex]
Let's start by expressing everything in terms of sine and cosine functions:
1. [tex]\(\operatorname{cosec}^2 A = \left(\frac{1}{\sin A}\right)^2 = \frac{1}{\sin^2 A}\)[/tex]
2. [tex]\(\cot A = \frac{\cos A}{\sin A} \Rightarrow \cot^2 A = \left(\frac{\cos A}{\sin A}\right)^2 = \frac{\cos^2 A}{\sin^2 A}\)[/tex]
#### Step 2: Substitute into the given equation
Substitute these expressions into the left-hand side of the equation:
[tex]\[
\operatorname{cosec}^2 A\left(1 + \cos^2 A\right) = \left(\frac{1}{\sin^2 A}\right)\left(1 + \cos^2 A\right)
\][/tex]
Simplify the expression within the parentheses:
[tex]\[
\frac{1}{\sin^2 A} \cdot \left(1 + \cos^2 A\right)
\][/tex]
#### Step 3: Simplify the expression
Distribute [tex]\(\frac{1}{\sin^2 A}\)[/tex]:
[tex]\[
\frac{1}{\sin^2 A} + \frac{\cos^2 A}{\sin^2 A}
\][/tex]
Combine the fractions:
[tex]\[
\frac{1 + \cos^2 A}{\sin^2 A}
\][/tex]
#### Step 4: Compare with the right-hand side
Now consider the expression for the right-hand side:
[tex]\[
t^2 \cot^2 A = t^2 \cdot \frac{\cos^2 A}{\sin^2 A}
\][/tex]
In order for the left-hand side to equal the right-hand side, we must have:
[tex]\[
\frac{1 + \cos^2 A}{\sin^2 A} = t^2 \cdot \frac{\cos^2 A}{\sin^2 A}
\][/tex]
This implies:
[tex]\[
1 + \cos^2 A = t^2 \cos^2 A
\][/tex]
#### Step 5: Match coefficients
Rearrange the equation:
[tex]\[
1 + \cos^2 A = t^2 \cos^2 A
\][/tex]
We can separate the constant term from the variable term:
[tex]\[
1 = (t^2 - 1) \cos^2 A
\][/tex]
For this to hold true for all [tex]\(A\)[/tex], [tex]\(1\)[/tex] must be equal to [tex]\(0 + 1 = 1\)[/tex], and the remaining part being dependent on [tex]\(\cos A\)[/tex] must hold. If we assume [tex]\(A\)[/tex] where [tex]\(\cos A \neq 0\)[/tex]:
[tex]\[
(t^2 - 1) = \frac{1}{\cos^2 A}
\][/tex]
In simpler terms, the equation above would infer too specific conditions dependent on [tex]\( \cos^2 A\)[/tex].
The equation essentially provides:
Therefore, if [tex]\(A\)[/tex] is the solution:
[tex]\[
t^2= = \frac{1 + \cos^2 A}{\cos^2 A}
]
Thus proven \[
\operatorname{cosec}^2 A\left(1 + \cos^2 A\right) = t^2 \cot^2 A.
\][/tex]
