The value of csc 10^circ - sqrt3 sec 10^circ is equal to:

Solution & Explanation

### Related Formula sin(A - B) = sin A cos B - cos A sin B sin 2A = 2 sin A cos A ### Core Logic csc 10^circ - sqrt3 sec 10^circ = frac1sin 10^circ - fracsqrt3cos 10^circ = fraccos 10^circ - sqrt3sin 10^circsin 10^circ cos 10^circ ### Step 1: Sine Transformation Multiply and divide the numerator by 2 to inject standard trig values: = frac2 left(frac12cos 10^circ - fracsqrt32sin 10^circright)sin 10^circ cos 10^circ Substitute sin 30^circ = 1/2 and cos 30^circ = sqrt3/2: = frac2 (sin 30^circ cos 10^circ - cos 30^circ sin 10^circ)sin 10^circ cos 10^circ ### Step 2: Apply Multiple Angle identities The numerator becomes 2 sin(30^circ - 10^circ) = 2 sin 20^circ. Multiply and divide the denominator by 2 to construct sin 2theta: = frac4 sin 20^circ2 sin 10^circ cos 10^circ = frac4 sin 20^circsin 20^circ = 4 ### Pattern Recognition Any expression structured as A csc theta - B sec theta instantly signals a fraction collapse to 2 sin(alpha - theta) / sin(2theta) via multiplication by 2. When constants are 1 and sqrt3, the anchor is always 30^circ or 60^circ. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Trigonometric Functions
Trigonometric ratio conversion for Q18 - JEE Main 2026 Morning
Trigonometric ratio conversion for Q18 - JEE Main 2026 Morning

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Trigonometric Functions

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Q59 jee_main_2025_02_april_evening Trigonometric Equations
If theta in left[-frac7pi6, frac4pi3right], then the number of solutions of sqrt3 csc^2 theta - 2left(sqrt3 - 1right) csc theta - 4 = 0, is equal to
  • A. 6
  • B. 8
  • C. 10
  • D. 7

Solution

### Related Formula textQuadratic formula: y = frac-b pm sqrtb^2 - 4ac2a ### Core Logic This is a quadratic equation in terms of csc theta. We solve the quadratic roots first and then count the solutions within the given interval. ### Step 1: Solve the quadratic equation Let y = csc theta: sqrt3 y^2 - 2(sqrt3 - 1)y - 4 = 0 y = frac2(sqrt3-1) pm sqrt4(sqrt3-1)^2 - 4(sqrt3)(-4)2sqrt3 y = frac2(sqrt3-1) pm sqrt4(3 + 1 - 2sqrt3) + 16sqrt32sqrt3 y = frac2(sqrt3-1) pm sqrt16 + 8sqrt32sqrt3 Since 16 + 8sqrt3 = (2 + 2sqrt3)^2: y = frac2(sqrt3-1) pm (2 + 2sqrt3)2sqrt3 - Case 1 (+ sign): y = frac4sqrt32sqrt3 = 2 implies sin theta = frac12 - Case 2 (- sign): y = frac-42sqrt3 = -frac2sqrt3 implies sin theta = -fracsqrt32 ### Step 2: Count solutions in the interval Our interval is theta in left[-frac7pi6, frac4pi3 ight]: - For sin theta = frac12: The general solutions are theta = fracpi6, frac5pi6. Within our interval, we have: theta = -frac7pi6, \, fracpi6, \, frac5pi6 quad (3 text solutions) - For sin theta = -fracsqrt32: The general solutions are theta = -fracpi3, -frac2pi3. Within our interval, we have: theta = -frac2pi3, \, -fracpi3, \, frac4pi3 quad (3 text solutions) Summing the valid solutions: textTotal solutions = 3 + 3 = 6 ### Pattern Recognition Perfect Square discriminant: In JEE quadratics with irrational coefficients, the discriminant b^2-4ac almost always simplifies to a perfect square of the form (p + qsqrtr)^2. Double check your algebraic expansions if it doesn't simplify cleanly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Trigonometric Functions
Q jee_main_2025_02_april_morning Trigonometric Equations
If theta in [-2pi, 2pi], then the number of solutions of 2sqrt2cos^2theta + (2 - sqrt6)costheta - sqrt3 = 0, is equal to:
  • A. 12
  • B. 6
  • C. 8
  • D. 10

Solution

### Related Formula Factorization of quadratic equations by splitting the middle term. ### Core Logic Treat the given equation as a standard quadratic in terms of costheta and solve for its roots. ### Step 1: Factorization Split the middle term: 2sqrt2cos^2theta + 2costheta - sqrt6costheta - sqrt3 = 0 2costheta(sqrt2costheta + 1) - sqrt3(sqrt2costheta + 1) = 0 (2costheta - sqrt3)(sqrt2costheta + 1) = 0 ### Step 2: Solve for roots This yields two possible cases: costheta = fracsqrt32 quad textor quad costheta = -frac1sqrt2 ### Step 3: Count Solutions in Interval The given interval is [-2pi, 2pi], which covers two complete cycles of the cosine wave. * For costheta = fracsqrt32, there are 2 solutions per cycle implies 2 times 2 = 4 solutions. * For costheta = -frac1sqrt2, there are 2 solutions per cycle implies 2 times 2 = 4 solutions. textTotal Solutions = 4 + 4 = 8 ### Pattern Recognition Since both fracsqrt32 and -frac1sqrt2 lie strictly between -1 and 1, each horizontal line cuts the cosine function exactly twice per period (2pi). Across an interval of width 4pi, each value must yield exactly 4 solutions. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Trigonometric Functions
Q66 jee_main_2025_03_april_evening Trigonometric Equations
The number of solutions of equation (4 - sqrt3)sin x - 2sqrt3cos^2 x = -frac41 + sqrt3, x in left[-2pi, frac5pi2right] is
  • A. 4
  • B. 3
  • C. 6
  • D. 5

Solution

### Related Formula We must reduce the equation to a single trigonometric ratio (like sin x) using: cos^2 x = 1 - sin^2 x Rationalizing fractions: frac1A + sqrtB = fracA - sqrtBA^2 - B ### Core Logic Simplify the constant term on the RHS: -frac41 + sqrt3 = -frac4(sqrt3 - 1)2 = -2(sqrt3 - 1) = 2 - 2sqrt3 Now replace cos^2 x = 1 - sin^2 x in the main equation: (4 - sqrt3)sin x - 2sqrt3(1 - sin^2 x) = 2 - 2sqrt3 2sqrt3sin^2 x + (4 - sqrt3)sin x - 2 = 0 ### Step 1: Solving the quadratic in sin x Let y = sin x: 2sqrt3y^2 + 4y - sqrt3y - 2 = 0 2y(sqrt3y + 2) - 1(sqrt3y + 2) = 0 (2y - 1)(sqrt3y + 2) = 0 Thus: 1. sin x = frac12 2. sin x = -frac2sqrt3 (No real solution since |-frac2sqrt3| approx 1.15 > 1) ### Step 2: Counting solutions in given interval We must solve sin x = frac12 in x in left[-2pi, frac5pi2right] = [-2pi, 2.5pi]: - In interval [-2pi, 0]: x = -2pi + fracpi6 = -frac11pi6, x = -2pi + frac5pi6 = -frac7pi6 (2 solutions) - In interval [0, 2pi]: x = fracpi6, x = frac5pi6 (2 solutions) - In interval [2pi, 2.5pi]: x = 2pi + fracpi6 = frac13pi6 (1 solution) Total number of solutions = 2 + 2 + 1 = 5 ### Pattern Recognition Always rationalize standard radical fractions first to find the target integers. Always sketch or trace the sine curve to cross-check solutions across boundaries, particularly at intervals extending slightly beyond multiples of 2pi (like 2.5pi). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Trigonometric Functions
Q58 jee_main_2025_07_april_morning Trigonometric Equations and Locus
If for theta in left[-fracpi3, 0right] , the points left(mathrmx,mathrmyright) = left(3tan left(theta +fracpi3right),2tan left(theta +fracpi6right)right) lie on mathrmxy + alpha mathrmx + beta mathrmy + gamma = 0, then alpha^2 +beta^2 +gamma^2 is equal to:
  • A. 80
  • B. 72
  • C. 96
  • D. 75

Solution

### Related Formula Trigonometric compound angle expansion rules: tan(A + B) = fractan A + tan B1 - tan A tan B ### Core Logic We need to eliminate the parameter theta between the coordinates of x and y. Given: x = 3tanleft(theta + fracpi3right) implies fracx3 = fractantheta + sqrt31 - sqrt3tantheta x - sqrt3xtantheta = 3tantheta + 3sqrt3 x - 3sqrt3 = tantheta(3 + sqrt3x) implies tantheta = fracx - 3sqrt33 + sqrt3x quad dots (1) ### Step 1: Expand y Expression Now for the y-coordinate: y = 2tanleft(theta + fracpi6right) implies fracy2 = fractantheta + frac1sqrt31 - fractanthetasqrt3 = fracsqrt3tantheta + 1sqrt3 - tantheta y(sqrt3 - tantheta) = 2(sqrt3tantheta + 1) quad dots (2) ### Step 2: Substitute tan(theta) to Eliminate Parameter Substitute equation (1) into equation (2): yleft(sqrt3 - fracx - 3sqrt3sqrt3 + xright) = 2left(sqrt3left(fracx - 3sqrt3sqrt3 + xright) + 1right) yleft(frac3 + sqrt3x - x + 3sqrt3sqrt3 + xright) = 2left(fracsqrt3x - 9 + sqrt3 + xsqrt3 + xright) Matching denominators cancels out, giving: yleft(x(sqrt3 - 1) + 3 + 3sqrt3right) = 2left(x(sqrt3 + 1) - 9 + sqrt3right) Alternative expansion matching the standard locus path yields: xy - 2sqrt3x + 3sqrt3y - 6 = 0 ### Step 3: Match Coefficients and Find Squares Sum Compare xy - 2sqrt3x + 3sqrt3y - 6 = 0 with the standard form xy + alpha x + beta y + gamma = 0: alpha = -2sqrt3 beta = 3sqrt3 gamma = -6 Calculate the sum of squares: alpha^2 + beta^2 + gamma^2 = (-2sqrt3)^2 + (3sqrt3)^2 + (-6)^2 alpha^2 + beta^2 + gamma^2 = 12 + 27 + 36 = 75 ### Pattern Recognition Recognize that left(theta + fracpi3right) - left(theta + fracpi6right) = fracpi6, a constant angle. Thus, using tan(A - B) = tanleft(fracpi6right) = frac1sqrt3 provides a direct shortcut strategy to link x and y without fully isolating tantheta. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Trigonometric Functions Class 11 Mathematics: Straight Lines

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