We are given two equations relating m and n with trigonometric terms secθ and tanθ. To prove a² + n² = b² + m², which can also be rearranged as m² - n² = a² - b², we will evaluate m² - n².

We are given two equations relating m and n with trigonometric terms secθ and tanθ. To prove a² + n² = b² + m², which can also be rearranged as m² - n² = a² - b², we will evaluate m² - n².

We square both given equations using the algebraic identity (x + y)² = x² + y² + 2xy. Notice that both expansions share the exact same cross-term, 2absec θ.

Subtracting n² from m², the identical cross-term 2absec θ cancels out. We then group the terms containing a² and b² together.

Since (b² - a²) = -(a² - b²), we factor out (a² - b²) to get (a² - b²)(sec²θ - tan²θ). Using the standard identity sec²θ - tan²θ = 1, this simplifies directly to a² - b².

Adding n² and b² to both sides of the equation gives a² + n² = b² + m², which completes the proof.

$(\sec\theta - \tan\theta) = \frac{1}{m}$ --- (2) Adding (1) & (2) $\sec\theta + \tan\theta + \sec\theta - \tan\theta = \frac{1}{m} + \frac{m}{1}$ $2\sec\theta = \frac{1 + m^2}{m}$ $\sec\theta = \frac{1 + m^2}{2m} \text{ (Hence proved)}$ 2) If $a\sec\theta + b\tan\theta = m$ , $b\sec\theta + a\tan\theta = n$ , prove that $a^2 + n^2 = b^2 + m^2$

Answer: Hence proved that $a^2 + n^2 = b^2 + m^2$ .

Step-by-step solution

Step 1: State the given equations

We are given two equations relating $m$ and $n$ with trigonometric terms $\sec\theta$ and $\tan\theta$ . To prove $a^2 + n^2 = b^2 + m^2$ , which can also be rearranged as $m^2 - n^2 = a^2 - b^2$ , we will evaluate $m^2 - n^2$ .

Step 2: Compute m squared and n squared

We square both given equations using the algebraic identity $(x + y)^2 = x^2 + y^2 + 2xy$ . Notice that both expansions share the exact same cross-term, $2ab\sec\theta\tan\theta$ .

Step 3: Subtract n squared from m squared

Subtracting $n^2$ from $m^2$ , the identical cross-term $2ab\sec\theta\tan\theta$ cancels out. We then group the terms containing $a^2$ and $b^2$ together.

Step 4: Apply trigonometric identity

Since $(b^2 - a^2) = -(a^2 - b^2)$ , we factor out $(a^2 - b^2)$ to get $(a^2 - b^2)(\sec^2\theta - \tan^2\theta)$ . Using the standard identity $\sec^2\theta - \tan^2\theta = 1$ , this simplifies directly to $a^2 - b^2$ .

Step 5: Rearrange to obtain the required identity

Adding $n^2$ and $b^2$ to both sides of the equation gives $a^2 + n^2 = b^2 + m^2$ , which completes the proof.

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