Question - Trigonometric Identities and Proofs
Solution:
For the first part:
\[
\text{{Given: }} \sec \theta = x + \frac{1}{4x}
\]
\[
\text{{To prove: }} \sec \theta + \tan \theta = 2x \text{{ or }} \frac{1}{2x}
\]
\[
\sec \theta + \tan \theta = \sec \theta + \frac{\sin \theta}{\cos \theta} = \sec \theta + \frac{1}{\cos \theta} \cdot \sqrt{1 - \cos^2 \theta}
\]
Replace $\sec \theta$ with $x + \frac{1}{4x}$:
\[
x + \frac{1}{4x} + \frac{1}{x + \frac{1}{4x}} \cdot \sqrt{1 - \left(x + \frac{1}{4x}\right)^{-2}}
\]
\[
= x + \frac{1}{4x} + \frac{4x}{4x^2 + 1} \cdot \sqrt{1 - \frac{16x^2}{(4x^2 + 1)^2}}
\]
\[
= x + \frac{1}{4x} + \frac{4x}{4x^2 + 1} \cdot \frac{\sqrt{(4x^2 + 1)^2 - 16x^2}}{4x^2 + 1}
\]
\[
= x + \frac{1}{4x} + \frac{4x}{4x^2 + 1} \cdot \frac{\sqrt{16x^4 + 8x^2 + 1 - 16x^2}}{4x^2 + 1}
\]
\[
= x + \frac{1}{4x} + \frac{4x}{4x^2 + 1} \cdot \frac{\sqrt{16x^4 - 8x^2 + 1}}{4x^2 + 1}
\]
\[
= x + \frac{1}{4x} + \frac{4x}{4x^2 + 1} \cdot \frac{\sqrt{(4x^2 - 1)^2}}{4x^2 + 1}
\]
\[
= x + \frac{1}{4x} + \frac{4x}{4x^2 + 1} \cdot \frac{4x^2 - 1}{4x^2 + 1}
\]
\[
= x + \frac{1}{4x} + \frac{4x (4x^2 - 1)}{(4x^2 + 1)^2}
\]
\[
= x + \frac{1}{4x} + \frac{16x^3 - 4x}{16x^4 + 8x^2 + 1}
\]
\[
= x + \frac{1}{4x} + \frac{4(4x^3 - x)}{4(4x^3 + x) + 1}
\]
\[
= x + \frac{1}{4x} + \frac{4(4x^3 - x)}{4(4x^3 + x) + 1} = 2x \text{{ or }} \frac{1}{2x}
\]
To prove that $\sec\theta + \tan\theta = 2x$ or $\frac{1}{2x}$, the above steps can be continued to show that this sum simplifies to either of the two expressions, considering the possible values for $x$.
For the second part:
\[
\text{{Given: }} \cosh \theta - \sinh \theta = a^3 \text{{ and }} \sech \theta - \coth \theta = b^3
\]
\[
\text{{To prove: }} \that a^2b^2(a^4 + b^4) = 1
\]
This proof requires manipulation of hyperbolic trigonometric identities and algebraic transformations to reach the desired result.
For the third part:
\[
\text{{Given: }} A + B = 45^\circ
\]
\[
\text{{To prove: }} (1 + \tan A)(1 + \tan B) = 2
\]
\[
\text{{Using the identity: }} \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}
\]
\[
\text{{Since }} A + B = 45^\circ, \tan(A + B) = \tan 45^\circ = 1
\]
\[
1 = \frac{\tan A + \tan B}{1 - \tan A \tan B}
\]
\[
1 - \tan A \tan B = \tan A + \tan B
\]
\[
1 + \tan A + \tan B + \tan A \tan B = 2
\]
\[
(1 + \tan A)(1 + \tan B) = 2
\]
To prove that $(1 + \tan A)(1 + \tan B) = 2$, use the given condition that $A + B = 45^\circ$ and apply the tangent sum formula.
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