Prove that (sinA+ cosecA)^2+(cosA+Sec2A)^2=7+tan^2A+Cot^2 A

To prove the given identity:

\[
(\sin A + \csc A)^2 + (\cos A + \sec A)^2 = 7 + \tan^2 A + \cot^2 A
\]

### Step 1: Expand the left-hand side (LHS)
We start by expanding each squared term on the left-hand side.

#### 1. Expanding \((\sin A + \csc A)^2\):
\[
(\sin A + \csc A)^2 = \sin^2 A + 2\sin A \cdot \csc A + \csc^2 A
\]
Since \(\csc A = \frac{1}{\sin A}\), we have:
\[
\sin A \cdot \csc A = 1
\]
Thus:
\[
(\sin A + \csc A)^2 = \sin^2 A + 2 + \csc^2 A
\]
Using \(\csc^2 A = \frac{1}{\sin^2 A}\), we get:
\[
(\sin A + \csc A)^2 = \sin^2 A + 2 + \frac{1}{\sin^2 A}
\]

#### 2. Expanding \((\cos A + \sec A)^2\):
\[
(\cos A + \sec A)^2 = \cos^2 A + 2\cos A \cdot \sec A + \sec^2 A
\]
Since \(\sec A = \frac{1}{\cos A}\), we have:
\[
\cos A \cdot \sec A = 1
\]
Thus:
\[
(\cos A + \sec A)^2 = \cos^2 A + 2 + \sec^2 A
\]
Using \(\sec^2 A = \frac{1}{\cos^2 A}\), we get:
\[
(\cos A + \sec A)^2 = \cos^2 A + 2 + \frac{1}{\cos^2 A}
\]

### Step 2: Add the expanded terms
Now, sum up both parts of the left-hand side:

\[
(\sin A + \csc A)^2 + (\cos A + \sec A)^2 = \left( \sin^2 A + 2 + \frac{1}{\sin^2 A} \right) + \left( \cos^2 A + 2 + \frac{1}{\cos^2 A} \right)
\]

Simplifying this expression:
\[
= \sin^2 A + \cos^2 A + 4 + \frac{1}{\sin^2 A} + \frac{1}{\cos^2 A}
\]
Since \(\sin^2 A + \cos^2 A = 1\), we have:
\[
1 + 4 + \frac{1}{\sin^2 A} + \frac{1}{\cos^2 A} = 5 + \frac{1}{\sin^2 A} + \frac{1}{\cos^2 A}
\]

### Step 3: Express \(\frac{1}{\sin^2 A} + \frac{1}{\cos^2 A}\) in terms of \(\tan A\) and \(\cot A\)
We know that:
\[
\frac{1}{\sin^2 A} = \csc^2 A = 1 + \cot^2 A
\]
and
\[
\frac{1}{\cos^2 A} = \sec^2 A = 1 + \tan^2 A
\]
Thus:
\[
\frac{1}{\sin^2 A} + \frac{1}{\cos^2 A} = (1 + \cot^2 A) + (1 + \tan^2 A) = 2 + \cot^2 A + \tan^2 A
\]

### Step 4: Final simplification
Substituting this into our previous expression:
\[
5 + \frac{1}{\sin^2 A} + \frac{1}{\cos^2 A} = 5 + 2 + \cot^2 A + \tan^2 A
\]
\[
= 7 + \cot^2 A + \tan^2 A
\]

Thus, the left-hand side simplifies to:

\[
(\sin A + \csc A)^2 + (\cos A + \sec A)^2 = 7 + \tan^2 A + \cot^2 A
\]

### Conclusion:
We have shown that:
\[
(\sin A + \csc A)^2 + (\cos A + \sec A)^2 = 7 + \tan^2 A + \cot^2 A
\]
The identity is proven to be true.

---

Would you like a more detailed explanation of any specific step or have any other questions?

Here are some related questions you might find useful:
1. Can you prove the Pythagorean identity \(\sin^2 A + \cos^2 A = 1\)?
2. How is \(\tan A\) related to \(\sin A\) and \(\cos A\)?
3. What are the properties of \(\sec A\) and \(\csc A\) in trigonometry?
4. Can you derive the double angle identities for sine and cosine?
5. How does the identity \(\tan^2 A + 1 = \sec^2 A\) relate to this proof?

**Tip:** When working with trigonometric identities, remember that many can be transformed into simpler forms using basic Pythagorean identities or reciprocal relationships between the trigonometric functions.

Prove that (sinA + cosecA)^2 + (cosA + Sec2A)^2 = 7 + tan^2A + Cot^2 A

This detailed proof demonstrates how to simplify and prove the trigonometric identity (sin A + csc A)^2 + (cos A + sec A)^2 = 7 + tan^2 A + cot^2 A. The solution involves key trigonometric identities such as the Pythagorean identity and reciprocal trigonometric functions. Suitable for students in Grades 11-12.

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