If \( h \sin ^{2} \theta_{1}+\cos ^{2} \theta_{1}=k, a_{1} \sin ^{2} \theta_{2}+k \tan ^{2} \theta_{2}=h \) \( 2 \sec ^{2} \theta_{2}=k \sin ^{2} \theta_{1}+1 \) and \( \sin \theta_{2}=\cos \theta_{2} \) find \( |b| \) when \( b^{2}=h-a \) and \( k=h-1 \).

To solve the problem, we start with the given equations and conditions. 1. We have the equations: \[ h \sin^2 \theta_1 + \cos^2 \theta_1 = k \] \[ a_1 \sin^2 \theta_2 + k \tan^2 \theta_2 = h \] \[ 2 \sec^2 \theta_2 = k \sin^2 \theta_1 + 1 \] \[ \sin \theta_2 = \cos \theta_2 \] 2. From the condition \(\sin \theta_2 = \cos \theta_2\), we can deduce that: \[ \tan \theta_2 = 1 \quad \Rightarrow \quad \theta_2 = 45^\circ \text{ or } \frac{\pi}{4} \] Therefore, we have: \[ \sin \theta_2 = \cos \theta_2 = \frac{\sqrt{2}}{2} \] 3. Now, substituting \(\sin^2 \theta_2\) and \(\tan^2 \theta_2\): \[ \sin^2 \theta_2 = \left(\frac{\sqrt{2}}{2}\right)^2 = \frac{1}{2} \] \[ \tan^2 \theta_2 = 1 \] 4. Substitute these values into the second equation: \[ a_1 \cdot \frac{1}{2} + k \cdot 1 = h \] This simplifies to: \[ \frac{a_1}{2} + k = h \quad \Rightarrow \quad a_1 = 2(h - k) \] 5. Now substitute \(\sin^2 \theta_1\) into the first equation. Let \(x = \sin^2 \theta_1\): \[ h x + \cos^2 \theta_1 = k \] Since \(\cos^2 \theta_1 = 1 - x\), we have: \[ h x + 1 - x = k \quad \Rightarrow \quad (h - 1)x + 1 = k \] 6. From the third equation: \[ 2 \sec^2 \theta_2 = k \sin^2 \theta_1 + 1 \] Since \(\sec^2 \theta_2 = 2\) (because \(\sec^2 45^\circ = 2\)): \[ 2 \cdot 2 = k \sin^2 \theta_1 + 1 \quad \Rightarrow \quad 4 = k \sin^2 \theta_1 + 1 \] This simplifies to: \[ k \sin^2 \theta_1 = 3 \quad \Rightarrow \quad \sin^2 \theta_1 = \frac{3}{k} \] 7. Now we have two expressions for \(k\): \[ k = (h - 1)x + 1 \] and \[ k = \frac{3}{\sin^2 \theta_1} \] 8. Setting these equal gives: \[ (h - 1)x + 1 = \frac{3}{x} \] Multiplying through by \(x\) (assuming \(x \neq 0\)): \[ (h - 1)x^2 + x = 3 \] Rearranging gives: \[ (h - 1)x^2 + x - 3 = 0 \] 9. Now we can use the quadratic formula to find \(x\): \[ x = \frac{-1 \pm \sqrt{1^2 - 4(h - 1)(-3)}}{2(h - 1)} = \frac{-1 \pm \sqrt{1 + 12(h - 1)}}{2(h - 1)} \] 10. We also know \(k = h - 1\). Thus, substituting \(k\) into \(b^2 = h - a\) and \(k = h - 1\): \[ b^2 = h - a \quad \text{and} \quad k = h - 1 \quad \Rightarrow