Question

Prove that : (sin A + sec A)\(^2\) + (cos A + cosec A)\(^2\) = (1 + sec A cosec A)\(^2\)

Hint:

An important algebraic-trigonometric shortcut to remember:
\[ \sec^2 A + \csc^2 A = \sec^2 A \csc^2 A \]
This identity converts a sum into a product and is extremely helpful in simplifying expressions containing sum of squared reciprocal functions.

Answer 1

Step 1: Rewrite each bracket on the left side as a single fraction, before squaring.
We want to prove:
\[ (\sin A + \sec A)^2 + (\cos A + \csc A)^2 = (1 + \sec A \csc A)^2 \]
Instead of expanding the squares right away, first write $\sec A = \frac{1}{\cos A}$ and $\csc A = \frac{1}{\sin A}$, and combine each bracket into one single fraction.
For the first bracket:
\[ \sin A + \sec A = \sin A + \frac{1}{\cos A} = \frac{\sin A \cos A + 1}{\cos A} \]
For the second bracket:
\[ \cos A + \csc A = \cos A + \frac{1}{\sin A} = \frac{\sin A \cos A + 1}{\sin A} \]
Both numerators come out identical, $\sin A \cos A + 1$, and this is the key simplification the whole route depends on.

Step 2: Square each single fraction and add them.
\[ \left(\frac{\sin A \cos A + 1}{\cos A}\right)^2 + \left(\frac{\sin A \cos A + 1}{\sin A}\right)^2 = (\sin A \cos A + 1)^2 \left(\frac{1}{\cos^2 A} + \frac{1}{\sin^2 A}\right) \]
Here the common squared factor $(\sin A \cos A + 1)^2$ was taken out since it appears in both pieces.

Step 3: Simplify the remaining bracket using the Pythagorean identity.
Combine the two fractions inside the bracket over a common denominator $\sin^2 A \cos^2 A$:
\[ \frac{1}{\cos^2 A} + \frac{1}{\sin^2 A} = \frac{\sin^2 A + \cos^2 A}{\sin^2 A \cos^2 A} \]
Using $\sin^2 A + \cos^2 A = 1$:
\[ = \frac{1}{\sin^2 A \cos^2 A} \]
So the full left side becomes:
\[ \text{LHS} = \frac{(\sin A \cos A + 1)^2}{\sin^2 A \cos^2 A} \]

Step 4: Write this as a single perfect square matching the right side.
Since $\sin^2 A \cos^2 A = (\sin A \cos A)^2$ is itself a perfect square:
\[ \text{LHS} = \left(\frac{\sin A \cos A + 1}{\sin A \cos A}\right)^2 = \left(1 + \frac{1}{\sin A \cos A}\right)^2 \]
Since $\frac{1}{\sin A \cos A} = \frac{1}{\cos A} \times \frac{1}{\sin A} = \sec A \csc A$:
\[ \text{LHS} = (1 + \sec A \csc A)^2 \]
This matches the right hand side exactly.

Final Answer:
By combining each bracket into a single fraction before squaring, instead of expanding the squares first, we reach the same identity, so $(\sin A + \sec A)^2 + (\cos A + \csc A)^2 = (1 + \sec A \csc A)^2$ is proved.