Question

Let \( \alpha = \tan \left( {{{5\pi } \over {16}}\sin \left( {2{{\cos }^{ - 1}}\left( {{1 \over {\sqrt 5 }}} \right)} \right)} \right) \) and \( \beta = \cos \left( {{{\sin }^{ - 1}}\left( {{4 \over 5}} \right) + {{\sec }^{ - 1}}\left( {{5 \over 3}} \right)} \right) \) where the inverse trigonometric functions take principal values. Then, the equation whose roots are \( \alpha \) and \( \beta \) is :

• A. \( 15{x^2} - 8x - 7 = 0 \)
• B. \( 5{x^2} - 12x + 7 = 0 \)
• C. \( 25{x^2} - 18x - 7 = 0 \)
• D. \( 25{x^2} - 32x + 7 = 0 \)

Hint:

Simplify \(\sin(2\cos^{-1}\tfrac1{\sqrt5})\) with the double-angle formula to get \(\alpha\); write both angles of \(\beta\) as right-triangle angles and use \(\cos(A+B)\).

Answer 1

The Correct Option is C

Step 1: Let \(\theta=\cos^{-1}\tfrac1{\sqrt5}\): \(\cos\theta=\tfrac1{\sqrt5}\), \(\sin\theta=\tfrac2{\sqrt5}\). \(\sin2\theta=2\cdot\tfrac2{\sqrt5}\cdot\tfrac1{\sqrt5}=\tfrac45\).
Step 2: \(\alpha=\tan\left(\tfrac{5\pi}{16}\cdot\tfrac45\right)=\tan\tfrac{\pi}{4}=1\).
Step 3: \(A=\sin^{-1}\tfrac45\): \(\sin A=\tfrac45,\cos A=\tfrac35\). \(B=\sec^{-1}\tfrac53\): \(\cos B=\tfrac35,\sin B=\tfrac45\).
Step 4: \(\beta=\cos(A+B)=\cos A\cos B-\sin A\sin B=\tfrac9{25}-\tfrac{16}{25}=-\tfrac7{25}\).
Step 5: Sum of roots \(=1-\tfrac7{25}=\tfrac{18}{25}\); product \(=-\tfrac7{25}\). Equation: \(x^2-\tfrac{18}{25}x-\tfrac7{25}=0\Rightarrow25x^2-18x-7=0\).
Answer: (C) \(25x^2-18x-7=0\)