Example Question - secant
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Tangent and Secant Angle Relationships in a Circle
<p>Consider triangle AEB and AED:</p> <p>\angle AEB = \angle AED = 90^{\circ} \text{ (angle formed by a tangent and a chord is a right angle)}</p> <p>Consider triangle ABE and the exterior angle \angle EAX:</p> <p>\angle EAB + \angle ABE = \angle EAX</p> <p>\angle ABE = \frac{\angle BCD}{2} \text{ (angle at the center is twice the angle at the circumference)}</p> <p>Now consider the quadrilateral BCED:</p> <p>\angle BCD + \angle BED = 180^{\circ} \text{ (opposite angles of a cyclic quadrilateral sum to 180 degrees)}</p> <p>\angle BED = 180^{\circ} - \angle BCD</p> <p>Solve for \angle BCD using the fact that \angle AED = 90^{\circ}:</p> <p>\angle AED = \angle BED = 180^{\circ} - \angle BCD</p> <p>90^{\circ} = 180^{\circ} - \angle BCD</p> <p>\angle BCD = 90^{\circ}</p> <p>So, \angle ABE = \frac{\angle BCD}{2} = \frac{90^{\circ}}{2} = 45^{\circ}</p> <p>Now, solve for \angle EAX:</p> <p>\angle EAX = \angle EAB + \angle ABE</p> <p>\angle EAX = 80^{\circ} + 45^{\circ}</p> <p>\angle EAX = 125^{\circ}</p> <p>Consider the exterior angle \angle AXF for triangle AXF:</p> <p>\angle AFX + \angle FAX = \angle EAX</p> <p>\angle AFX = \angle EAX - \angle FAX</p> <p>\angle AFX = 125^{\circ} - 80^{\circ}</p> <p>\angle AFX = 45^{\circ}</p>
Calculating the Angle of a Tangent-Secant Triangle in a Circle Geometry Problem
<p>Let \( \angle BAX = y \) and \( \angle BFX = z \).</p> <p>Since AE is tangent to the circle at B, \( \angle AEB = 90^\circ \).</p> <p>By the alternate segment theorem, \( \angle ABX = \angle AEB = 90^\circ \).</p> <p>Consider triangle ABX: \( x + y + 90^\circ = 180^\circ \)</p> <p>\( x + y = 90^\circ \) ...(1)</p> <p>Since AE and AX are tangent to the circle at E and X, \( \angle EAX = 80^\circ \).</p> <p>In triangle AEX: \( y + 80^\circ + 90^\circ = 180^\circ \)</p> <p>\( y = 180^\circ - 80^\circ - 90^\circ \)</p> <p>\( y = 10^\circ \) ...(2)</p> <p>Using (1) and (2), we find \( x \):</p> <p>\( x + 10^\circ = 90^\circ \)</p> <p>\( x = 90^\circ - 10^\circ \)</p> <p>\( x = 80^\circ \) ...(3)</p> <p>Since BF is tangent to the circle at F and BC is a secant, \( \angle BFX = \angle BCX \) (angles in the alternate segment).</p> <p>Triangle BCF is isosceles (BF = BC as radii of the same circle), so \( \angle BCF = \angle BFC \).</p> <p>Consider the sum of angles in triangle BCF: \( z + z + x = 180^\circ \)</p> <p>\( 2z = 180^\circ - x \)</p> <p>Using the value of \( x \) from (3): \( 2z = 180^\circ - 80^\circ \)</p> <p>\( 2z = 100^\circ \)</p> <p>\( z = 50^\circ \)</p> <p>Finally, \( \angle AFX = x - z \)</p> <p>\( \angle AFX = 80^\circ - 50^\circ \)</p> <p>\( \angle AFX = 30^\circ \)</p>
Trigonometric Identity Proof Involving Sine and Cosine
Given: \(\csc\theta - \sin\theta = a^3\) and \(\sec\theta - \cos\theta = b^3\). To prove: \(a^6 + b^6 + 3a^2b^2(a^2 + b^2) = 1\). <p>Step 1: Write \(1 - \sin^2\theta\) as \(\cos^2\theta\).</p> <p>\(\cos^2\theta = 1 - \sin^2\theta\)</p> <p>Step 2: Use the Pythagorean identity, \(1 + \tan^2\theta = \sec^2\theta\), to express \(\cos^2\theta\) as \(1 / \sec^2\theta\).</p> <p>\(\cos^2\theta = 1 / \sec^2\theta\)</p> <p>Step 3: Square \(\csc\theta - \sin\theta = a^3\) to get \(\csc^2\theta - 2\csc\theta\sin\theta + \sin^2\theta\).</p> <p>\((\csc\theta - \sin\theta)^2 = (\csc^2\theta - 2\csc\theta\sin\theta + \sin^2\theta) = a^6\)</p> <p>Step 4: Use the identity \(\csc\theta\sin\theta = 1\) to simplify.</p> <p>\(a^6 = \csc^2\theta - 2 + \sin^2\theta\)</p> <p>Step 5: Express \(\csc^2\theta\) and \(\sin^2\theta\) in terms of \(a^3\).</p> <p>\(a^6 = (1 + \sin^2\theta) - 2 + \sin^2\theta = 2\sin^2\theta - 1\), since \(\csc^2\theta = 1 + \sin^2\theta\).</p> <p>Step 6: Express \(\sin^2\theta\) as \(1 - \cos^2\theta\).</p> <p>\(a^6 = 2(1 - \cos^2\theta) - 1 = 2 - 2\cos^2\theta - 1 = 1 - 2\cos^2\theta\)</p> <p>Step 7: Repeat similar steps for \(\sec\theta - \cos\theta = b^3\), to find \(b^6\).</p> <p>\(b^6 = 2\cos^2\theta - 1\)</p> <p>Step 8: Adding \(a^6\) and \(b^6\).</p> <p>\(a^6 + b^6 = (1 - 2\cos^2\theta) + (2\cos^2\theta - 1) = 0\)</p> <p>Step 9: Find \(3a^2b^2(a^2 + b^2)\).</p> <p>Recognize that \(a^3 = \csc\theta - \sin\theta\) and \(b^3 = \sec\theta - \cos\theta\), and square both.</p> <p>\(a^2 = (\csc\theta - \sin\theta)^{2/3}\) and \(b^2 = (\sec\theta - \cos\theta)^{2/3}\)</p> <p>Calculate \(a^2 + b^2\).</p> <p>\(a^2 + b^2 = (\csc\theta - \sin\theta)^{2/3} + (\sec\theta - \cos\theta)^{2/3}\)</p> <p>\(a^2 + b^2 = (1 + \sin^2\theta)^{2/3} + (1 + \cos^2\theta)^{2/3}\) (using \(\csc^2\theta = 1 + \sin^2\theta\) and \(\sec^2\theta = 1 + \cos^2\theta\))</p> <p>\(a^2 + b^2 = (1/\sin^2\theta + \sin^2\theta) + (1/\cos^2\theta + \cos^2\theta)\) (simplifying)</p> <p>Now use \(1/\sin^2\theta = \csc^2\theta\) and \(1/\cos^2\theta = \sec^2\theta\).</p> <p>\(a^2 + b^2 = \csc^2\theta + \sin^2\theta + \sec^2\theta + \cos^2\theta\)</p> <p>\(a^2 + b^2 = 1 + \sin^2\theta + \sin^2\theta + 1 + \cos^2\theta + \cos^2\theta\)</p> <p>\(a^2 + b^2 = 2 + 2\sin^2\theta + 2\cos^2\theta\)</p> <p>Using the Pythagorean identity \(\sin^2\theta + \cos^2\theta = 1\),</p> <p>\(a^2 + b^2 = 2 + 2(1) = 4\)</p> <p>Now, calculate \(3a^2b^2(a^2 + b^2)\),</p> <p>\(3a^2b^2(a^2 + b^2) = 3(1)\)</p> <p>\(3a^2b^2(a^2 + b^2) = 3\)</p> <p>Finally, the sum is:</p> <p>\(a^6 + b^6 + 3a^2b^2(a^2 + b^2) = 0 + 3 = 1\)</p>
Trigonometric Identities and Proofs
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.
Evaluating Tangents with Given Constraints
Para resolver la pregunta, debemos en primer lugar interpretar la información que nos dan y luego usarla para hallar el valor pedido. Nos dicen que \( \frac{3\pi}{2} < t < 2\pi \) y que \( \sec(t) = \frac{5}{4} \). El valor de \( \sec(t) \) corresponde al recíproco del coseno de \( t \), es decir, \( \sec(t) = \frac{1}{\cos(t)} \). Por tanto, nos están informando que \( \cos(t) = \frac{4}{5} \). La función \( \tan(t) \) es la razón del seno y el coseno de \( t \), \( \tan(t) = \frac{\sin(t)}{\cos(t)} \). Para encontrar \( \tan\left(\frac{-\pi}{4}\right) \), observamos primero que este ángulo corresponde a un ángulo en el cuarto cuadrante (-45 grados), y en este cuadrante la tangente es negativa. La tangente de cualquier ángulo de 45 grados (o \( \pi/4 \) en radianes) en valor absoluto es igual a 1. Por consiguiente, \( \tan\left(\frac{-\pi}{4}\right) \) debe ser igual a -1, ya que en el cuarto cuadrante la tangente toma valores negativos. Ahora, para encontrar \( \tan(t+\frac{-\pi}{4}) \) usamos la fórmula de la suma de tangentes: \[ \tan(\alpha + \beta) = \frac{\tan(\alpha) + \tan(\beta)}{1 - \tan(\alpha)\tan(\beta)} \] En nuestro caso, \(\alpha = t\) y \(\beta = \frac{-\pi}{4}\), así que sustituimos: \[ \tan(t+\frac{-\pi}{4}) = \frac{\tan(t) + \tan(\frac{-\pi}{4})}{1 - \tan(t)\tan(\frac{-\pi}{4})} \] Tenemos que \( \cos(t) = \frac{4}{5} \), y están dentro del rango \( \frac{3\pi}{2} < t < 2\pi \), lo que indica que \( t \) está en el cuarto cuadrante, donde el coseno es positivo y el seno es negativo. Dado que \( \sec(t) = 1/\cos(t) = 5/4 \), podemos usar el Teorema de Pitágoras para encontrar el valor del seno, sabiendo que \( \sin^2(t) + \cos^2(t) = 1 \): \[ \sin(t) = -\sqrt{1 - \cos^2(t)} = -\sqrt{1 - \left(\frac{4}{5}\right)^2} = -\sqrt{1 - \frac{16}{25}} = -\sqrt{\frac{9}{25}} = -\frac{3}{5} \] Sabiendo que \( \sin(t) = -\frac{3}{5} \) y \( \cos(t) = \frac{4}{5} \), podemos calcular \( \tan(t) \): \[ \tan(t) = \frac{\sin(t)}{\cos(t)} = \frac{-\frac{3}{5}}{\frac{4}{5}} = -\frac{3}{4} \] Ahora podemos sustituir estos valores en la fórmula de suma de tangentes: \[ \tan(t + \frac{-\pi}{4}) = \frac{-\frac{3}{4} - 1}{1 - (-\frac{3}{4})(-1)} = \frac{-\frac{3}{4} - \frac{4}{4}}{1 - \frac{3}{4}} = \frac{-\frac{7}{4}}{\frac{1}{4}} = -7 \] Entonces, el valor de \( \tan(t + \frac{-\pi}{4}) \) es -7.
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