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MathsMediumMCQ2013 · 09 Apr Online

Q67.A value of x for which sin (cot−1(1 + x)) = cos (tan−1 x), is : (1) −12 (2) 1 (3) 0 (4) 1 2

What This Question Tests

This question requires converting inverse trigonometric functions to their algebraic forms using right-angled triangles or identities and then solving the resulting algebraic equation for x.

Concepts Tested

Inverse trigonometric identitiesSolving trigonometric equations

Formulas Used

sin(cot⁻¹x) = 1/√(1+x²)

cos(tan⁻¹x) = 1/√(1+x²) (for x>0) or x/√(1+x^2)

Identities related to inverse trig functions

📚 NCERT Sections This Tests

9.15Apply Mirror Equation And The Condition:

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9.17(A) Sin I¢C = 1.44/1.68 Which Gives I¢C = 59°. Total Internal Reflection

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9.17 (a) sin i¢c = 1.44/1.68 which gives i¢c = 59°. Total internal reflection takes place when i > 59° or when r < rmax = 31°. Now, (sin i /sin r max max ) = 1.68 , which gives imax ~ 60°. Thus, all incident rays of angles in the range 0 < i < 60° will suffer total internal reflections in the pipe. (If the length of the pipe is finite, which it is in practice, there will be a lower limit on i determined by the ratio of the diameter to the length of the pipe.) (b) If there is no outer coating, i¢c = sin–1(1/1.68) = 36.5°. Now, i = 90° will have r = 36.5° and i¢ = 53.5° which is greater than i¢c. Thus, all incident rays (in the range 53.5° < i < 90°) will suffer total internal reflections.