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

Reference Study Guides

Trigonometric Functions

More Trigonometric Functions Previous-Year Questions — Page 2

Q52 jee_main_2025_29_jan_evening Trigonometric Identities
If sin x + sin^2 x = 1, x in left(0, fracpi2right), then (cos^12 x + tan^12 x) + 3 (cos^10 x + tan^10 x + cos^8 x + tan^8 x) + (cos^6 x + tan^6 x) is equal to
  • A. 4
  • B. 3
  • C. 2
  • D. 1

Solution

### Related Formula Fundamental identities: sin^2 x + cos^2 x = 1 tan x = fracsin xcos x Algebraic identity for a perfect cube: (A + B)^3 = A^3 + 3A^2B + 3AB^2 + B^3 ### Core Logic Given equation: sin x + sin^2 x = 1 implies sin x = 1 - sin^2 x = cos^2 x Dividing both sides by cos^2 x: fracsin xcos^2 x = 1 implies tan x sec x = 1 implies tan x = cos x ### Step 1: Simplify the Expression Since tan x = cos x, we can substitute tan x with cos x throughout the given expression: (cos^12 x + cos^12 x) + 3(cos^10 x + cos^10 x + cos^8 x + cos^8 x) + (cos^6 x + cos^6 x) = 2cos^12 x + 6cos^10 x + 6cos^8 x + 2cos^6 x = 2left[cos^12 x + 3cos^10 x + 3cos^8 x + cos^6 xright] ### Step 2: Apply the Cubic Identity Notice that the expression inside the brackets matches the expansion of a perfect cube: = 2left[(cos^4 x + cos^2 x)^3right] Since cos^2 x = sin x, it follows that cos^4 x = sin^2 x. Substituting these back in: = 2left[(sin^2 x + sin x)^3right] We know from the problem statement that sin x + sin^2 x = 1. Therefore: = 2(1)^3 = 2 ### Pattern Recognition When given sin x + sin^2 x = 1, the substitution cos^2 x = sin x or tan x = cos x is a classic identity trick. Recognizing binomial coefficients (1, 3, 3, 1) immediately signals to condense into a full cube structure. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Trigonometric Functions
Q60 jee_main_2025_03_april_morning Trigonometric Equations
The number of solutions of the equation 2x + 3tan x = pi, xin[-2pi, 2pi] - left\pmfracpi2, pmfrac3pi2right\ is[cite: 612, 613]:
  • A. 6
  • B. 5
  • C. 4
  • D. 3

Solution

### Related Formula Intersection method for transcendental configurations: f(x) = g(x) Plot both curves separately to observe distinct intersection markers inside target domain spans.
Trigonometric Equations diagram for Q60 - JEE Main 2025 Morning
Trigonometric Equations diagram for Q60 - JEE Main 2025 Morning
### Core Logic Rearrange terms to group equations into known standard graphing profiles [cite: 1321]: 3tan x = pi - 2x implies tan x = fracpi3 - frac2x3 [cite: 1321] Plot the linear equation line y = fracpi3 - frac2x3 along with multiple period tracks of the trigonometric function y = tan x within the interval [-2pi, 2pi][cite: 613, 1321]. ### Step 1: Point analysis across branches The line has a negative slope and passes through (0, pi/3) and (3pi/2, 0). Looking across distinct interval chunks separated by asymptotes[cite: 613]: 1. Branch 1 left(-2pi, -frac3pi2right): 1 intersection 2. Branch 2 left(-frac3pi2, -fracpi2right): 1 intersection 3. Branch 3 left(-fracpi2, fracpi2right): 1 intersection near origin 4. Branch 4 left(fracpi2, frac3pi2right): 1 intersection 5. Branch 5 left(frac3pi2, 2piright): 1 intersection Counting all distinct points across valid domains gives 5 solutions total[cite: 1337]. ### Pattern Recognition A linear curve intersecting tangent asymptote branches will cut exactly once through every continuous range slice unless the line is strictly horizontal or parallel to asymptotes. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Trigonometric Functions
Q70 jee_main_2025_04_april_morning Trigonometric Identities
If 10sin^4theta + 15cos^4theta = 6, then the value of frac27csc^6theta + 8sec^6theta16sec^8theta is:
  • A. frac25
  • B. frac34
  • C. frac35
  • D. frac15

Solution

### Related Formula Trigonometric identity conversion: cos^2theta = 1 - sin^2theta ### Core Logic Let sin^2theta = t. Substitute this into the given equation: 10t^2 + 15(1 - t)^2 = 6 implies 10t^2 + 15(1 - 2t + t^2) = 6 25t^2 - 30t + 9 = 0 implies (5t - 3)^2 = 0 implies t = frac35 Thus, sin^2theta = frac35 and cos^2theta = frac25. ### Step 1: Simplify Target Expression Find individual terms from inverse relations: csc^2theta = frac53 implies csc^6theta = frac12527 sec^2theta = frac52 implies sec^6theta = frac1258 sec^8theta = left(frac52right)^4 = frac62516 Substitute values into expression: textNumerator = 27left(frac12527right) + 8left(frac1258right) = 125 + 125 = 250 textDenominator = 16left(frac62516right) = 625 ### Step 2: Conclusion textValue = frac250625 = frac25 ### Pattern Recognition Equations structured as Asin^4theta + Bcos^4theta = C often yield perfect square trinomial combinations. Check for clean coefficient cancelation steps before computing higher power expressions. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Trigonometric Functions
Q59 jee_main_2025_07_april_evening Trigonometric Equations
The number of solutions of the equation cos 2 theta cos frac theta2 + cos frac 5 theta2 = 2 cos^ 3 frac 5 theta2 in left[ - frac pi2, frac pi2 right] is:
  • A. 7
  • B. 5
  • C. 6
  • D. 9

Solution

### Related Formula Product-to-sum formula and triple angle identity are: 2cos Acos B = cos(A+B) + cos(A-B) 2cos^3 theta = frac12(cos 3theta + 3cos theta) ### Core Logic Given equation: cos 2 theta cos frac theta2 + cos frac 5 theta2 = 2 cos^ 3 frac 5 theta2 Multiplying by 2: 2cos 2theta cos fractheta2 + 2cos frac5theta2 = 4cos^3 frac5theta2 Using product-to-sum on the first term: left(cosfrac5theta2 + cosfrac3theta2right) + 2cos frac5theta2 = 2 left(cos frac15theta2 + 3cos frac5theta2right) cosfrac3theta2 + 3cosfrac5theta2 = 2cosfrac15theta2 + 6cosfrac5theta2 cosfrac3theta2 - 3cosfrac5theta2 = 2cosfrac15theta2 ### Step 1: Structural Rearrangement Simplifying through standard trigonometric transformation equations leads directly to: cosfrac3theta2 = cosfrac15theta2 cosfrac15theta2 - cosfrac3theta2 = 0 2sin(3theta)sinleft(frac9theta2right) = 0 Hence, either sin(3theta) = 0 or \sin\left(\frac{9\theta}{2}\right) = 0. ### Step 2: Finding Roots in the Interval Interval given: \theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]. Case A: sin(3\theta) = 0 \implies 3\theta = n\pi \implies \theta = \frac{n\pi}{3} Values inside interval: \left\{-\frac{pi}{3}, 0, \frac{\pi}{3}\right\} (3 solutions). Case B: sin\left(\frac{9\theta}{2}\right) = 0 \implies \frac{9\theta}{2} = m\pi \implies \theta = \frac{2m\pi}{9} Values inside interval: \left\{-\frac{4\pi}{9}, -\frac{2\pi}{9}, 0, \frac{2\pi}{9}, \frac{4\pi}{9}\right\}. Since 0 is already counted, this gives 4 unique additional solutions. Total unique solutions = 3 + 4 = 7. ### Pattern Recognition Transforming powers like \cos^3 x$ back into simple multiple-angle terms linearizes trigonometric equations instantly for direct factoring. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Trigonometry
Q55 jee_main_2025_24_jan_evening Trigonometric Equations and Solutions
Let A=left\xin(0,pi)-left\fracpi2right\:log_(2/pi)|sin x|+log_(2/pi)|cos x|=2right\ and B=left\xge0:sqrtx(sqrtx-4)-3|sqrtx-2|+6=0right\. Then n(Acup B) is equal to:
  • A. 4
  • B. 2
  • C. 8
  • D. 6

Solution

### Related Formula Logarithmic addition property: log(a) + log(b) = log(ab). Double \angle sine formula: 2sin x cos x = sin 2x. ### Step 1: Simplify Set A Combine the logarithmic elements: log_(2/pi) left(|sin x| cdot |cos x|right) = 2 |sin x cos x| = left(frac2piright)^2 = frac4pi^2 |2sin x cos x| = frac8pi^2 Rightarrow |sin 2x| = frac8pi^2 Since pi^2 approx 9.87, frac8pi^2 approx 0.81, which is less than 1. Plotting |sin 2x| = frac8pi^2 over the specified range x in (0, pi) yields exactly 4 real intersection points[cite: 3314, 3906].
Trigonometric Equations graph for Q55 - JEE Main 2025 Evening
Trigonometric Equations graph for Q55 - JEE Main 2025 Evening
Hence, n(A) = 4. ### Step 2: Simplify Set B Let sqrtx = t where t ge 0. The equation becomes[cite: 3316, 3910]: t(t-4) - 3|t-2| + 6 = 0 **Case I:** If t < 2 : t^2 - 4t - 3(-(t-2)) + 6 = 0 Rightarrow t^2 - 4t + 3t - 6 + 6 = 0 Rightarrow t^2 - t = 0 t = 0, 1 Rightarrow x = 0, 1 [cite: 3912, 3913] **Case II:** If t > 2 : t^2 - 4t - 3(t-2) + 6 = 0 Rightarrow t^2 - 4t - 3t + 6 + 6 = 0 Rightarrow t^2 - 7t + 12 = 0 [cite: 3915, 3916] (t-3)(t-4) = 0 Rightarrow t = 3, 4 Rightarrow x = 9, 16 [cite: 3917, 3918] Hence, set B = \0, 1, 9, 16\, giving n(B) = 4. ### Step 3: Calculate Union Since all elements of set A are non-integral angles in (0, pi) and elements of set B are pure integers, the sets are completely disjoint (A cap B = emptyset). n(A cup B) = n(A) + n(B) = 4 + 4 = 8 ### Pattern Recognition Always separate functions into disjoint numeric domains (e.g., angles vs. whole integers) to conclude unions without performing tedious tracking of individual values manually. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Trigonometric Functions Class 11 Mathematics: Sets

More Trigonometric Functions Questions — jee_main_2026_21_jan_morning

Practice all Trigonometric Functions previous-year questions →

Rankbit System
JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)
YOUR FIRST PREP STEP STARTS HERE

We Map Every Repeating Question in Competitive Exams.

Say goodbye to generic mock test fatigue. RankBit uses smart analysis to group past exam questions into their foundational Repeating Question Types. Find chapter weightage, track repeating questions, and score higher with targeted practice.

Select Your Target Exam

Choose an exam track below to find formulas per chapter and patterns.

Syncing Exam Intelligence

Mapping formulas and patterns across all tracks…

PATH A — FULL LENGTH PRACTICE

Full Mock Test Hub

Simulate real NTA exam conditions with fully tracked mocks. Time yourself against past papers.

Now Live Open
PATH B — TARGETED PRACTICE

Topic-wise Practice Hub

Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.

Loading Questions... Browse Topics

Exam Analysis Tools

Chapter weightage trends and topic-wise practice — all mapped from past shift papers.

Year-wise PYQ

Chapter Weightage & Trends

See high-yield chapters, year-wise counts and the frequency-based chance each topic repeats next exam.

Weightage Board Open
Targeted PYQs

Topic-wise Practice Hub

Every past-year question for a chapter pulled from all papers, with its key formulas alongside.

One chapter at a time Open
Latest from the Blog
View all →

Loading articles...