🏆 Daily Math Board Challenge (2026 Edition)
Check whether (x + 1)² = 2(x − 3) is a quadratic equation.
🎯 Chapter Readiness Score35%
⚡ Key Formulas at a Glance
Formula (Hover/Click)
Complementary Sum 1
sin⁻¹x + cos⁻¹x = π/2 for x ∈ [−1, 1]
Formula (Hover/Click)
Complementary Sum 2
tan⁻¹x + cot⁻¹x = π/2 for x ∈ ℜ
Formula (Hover/Click)
Addition Identity
tan⁻¹x + tan⁻¹y = tan⁻¹[(x+y) / (1−xy)] for xy < 1
Formula (Hover/Click)
2tan⁻¹x Identities
2tan⁻¹x = sin⁻¹(2x/(1+x²)) = cos⁻¹((1−x²)/(1+x²))
🔍
1NCERT Ex 2.1 Q11 • 2 Marks2 Marks
Find the principal value of cos⁻¹(-1/2) + 2 sin⁻¹(1/2).
▶ Show Detailed Solution
Step 1 (Principal Values):
cos⁻¹(-1/2) = π - π/3 = 2π/3.
sin⁻¹(1/2) = π/6.
Step 2 (Evaluate):
Expression = 2π/3 + 2(π/6) = 2π/3 + π/3 = 3π/3 = π.
✔ Final Answer: π
2NCERT Ex 2.1 Q12 • 2 Marks2 Marks
Evaluate tan⁻¹(1) + cos⁻¹(-1/2) + sin⁻¹(-1/2).
▶ Show Detailed Solution
Step 1 (Individual Values):
tan⁻¹(1) = π/4
cos⁻¹(-1/2) = 2π/3
sin⁻¹(-1/2) = -π/6
Step 2 (Sum):
π/4 + 2π/3 - π/6 = (3π + 8π - 2π)/12 = 3π/4.
✔ Final Answer: 3π/4
3CBSE Board 2024 • 3 Marks3 Marks
Solve for x: tan⁻¹(2x) + tan⁻¹(3x) = π/4.
▶ Show Detailed Solution
Step 1 (Formula Application):
tan⁻¹[(2x + 3x)/(1 - 6x²)] = π/4.
Step 2 (Algebraic Solution):
5x / (1 - 6x²) = tan(π/4) = 1
5x = 1 - 6x² ⟹ 6x² + 5x - 1 = 0
(6x - 1)(x + 1) = 0 ⟹ x = 1/6 or x = -1.
Since x = -1 makes LHS negative, x = 1/6.
✔ Final Answer: x = 1/6
4NCERT Ex 2.2 Q3 • 3 Marks3 Marks
Prove that tan⁻¹(1/2) + tan⁻¹(2/11) = tan⁻¹(3/4).
▶ Show Detailed Solution
Step 1 (LHS Expansion):
LHS = tan⁻¹[ (1/2 + 2/11) / (1 - (1/2)(2/11)) ]
Step 2 (Simplify Fraction):
= tan⁻¹[ (15/22) / (20/22) ] = tan⁻¹(15/20) = tan⁻¹(3/4) = RHS. ✓
✔ Hence Proved.
5NCERT Ex 2.2 Q5 • 4 Marks4 Marks
Simplify tan⁻¹[ (√(1 + x²) - 1) / x ], x ≠ 0.
▶ Show Detailed Solution
Step 1 (Substitution):
Let x = tan θ ⟹ θ = tan⁻¹ x.
Step 2 (Trigonometric Identity):
√(1 + tan² θ) - 1 = sec θ - 1 = (1 - cos θ)/sin θ = tan(θ/2).
tan⁻¹(tan(θ/2)) = θ/2 = (1/2) tan⁻¹ x.
✔ Final Answer: (1/2) tan⁻¹ x
6NCERT Ex 2.2 Q8 • 3 Marks3 Marks
Express tan⁻¹[ (cos x - sin x)/(cos x + sin x) ], -π/4 < x < 3π/4 in simplest form.
▶ Show Detailed Solution
Step 1 (Divide by cos x):
tan⁻¹[ (1 - tan x)/(1 + tan x) ]
Step 2 (Identity):
= tan⁻¹[ tan(π/4 - x) ] = π/4 - x.
✔ Final Answer: π/4 - x
7CBSE Board 2023 • 4 Marks4 Marks
Prove that sin⁻¹(3/5) - sin⁻¹(8/17) = cos⁻¹(84/85).
▶ Show Detailed Solution
Step 1 (Convert to tan⁻¹):
Let sin⁻¹(3/5) = A ⟹ tan A = 3/4.
Let sin⁻¹(8/17) = B ⟹ tan B = 8/15.
Step 2 (tan(A - B)):
tan(A - B) = (3/4 - 8/15)/(1 + 24/60) = (13/60)/(84/60) = 13/84.
cos(A - B) = 84/85 ⟹ A - B = cos⁻¹(84/85) = RHS.
✔ Hence Proved.
8CBSE Board 2023 HOTS • 5 Marks5 Marks
Solve for x: sin⁻¹(1 - x) - 2 sin⁻¹ x = π/2.
▶ Show Detailed Solution
Step 1 (Rearrange):
sin⁻¹(1 - x) = π/2 + 2 sin⁻¹ x
1 - x = sin(π/2 + 2 sin⁻¹ x) = cos(2 sin⁻¹ x).
Step 2 (Substitute x = sin θ):
1 - sin θ = cos 2θ = 1 - 2 sin² θ ⟹ 2 sin² θ - sin θ = 0
sin θ (2 sin θ - 1) = 0 ⟹ x = 0 or x = 1/2.
Testing x = 1/2 gives LHS = sin⁻¹(1/2) - 2 sin⁻¹(1/2) = -π/6 ≠ π/2.
Testing x = 0 gives LHS = sin⁻¹(1) - 0 = π/2 ✓.
✔ Final Answer: x = 0
9NCERT Ex 2.2 Q9 • 4 Marks4 Marks
Find the value of tan( (1/2)[sin⁻¹(2x/(1+x²)) + cos⁻¹((1-y²)/(1+y²))] ), |x| < 1, y > 0.
▶ Show Detailed Solution
Step 1 (Use 2 tan⁻¹ identities):
sin⁻¹(2x/(1+x²)) = 2 tan⁻¹ x
cos⁻¹((1-y²)/(1+y²)) = 2 tan⁻¹ y
Step 2 (Simplify Expression):
tan[ (1/2)(2 tan⁻¹ x + 2 tan⁻¹ y) ] = tan[ tan⁻¹ x + tan⁻¹ y ] = (x + y)/(1 - xy).
✔ Final Answer: (x + y)/(1 - xy)
10CBSE Board 2022 • 4 Marks4 Marks
Prove that 2 tan⁻¹(1/3) + tan⁻¹(1/7) = π/4.
▶ Show Detailed Solution
Step 1 (2 tan⁻¹(1/3)):
2 tan⁻¹(1/3) = tan⁻¹[ (2/3)/(1 - 1/9) ] = tan⁻¹(3/4).
Step 2 (Add tan⁻¹(1/7)):
tan⁻¹(3/4) + tan⁻¹(1/7) = tan⁻¹[ (3/4 + 1/7)/(1 - 3/28) ] = tan⁻¹(25/25) = tan⁻¹(1) = π/4.
✔ Hence Proved.
11NCERT Ex 2.1 Q11 (Variant #11) • 2 Marks2 Marks
Find the principal value of cos⁻¹(-3/4) + 2 sin⁻¹(3/4).
▶ Show Detailed Solution
Step 1 (Principal Values):
cos⁻¹(-1/2) = π - π/3 = 2π/3.
sin⁻¹(1/2) = π/6.
Step 2 (Evaluate):
Expression = 2π/3 + 2(π/6) = 2π/3 + π/3 = 3π/3 = π.
✔ Final Answer: π
12NCERT Ex 2.1 Q12 (Variant #12) • 2 Marks2 Marks
Evaluate tan⁻¹(1) + cos⁻¹(-1/2) + sin⁻¹(-1/2).
▶ Show Detailed Solution
Step 1 (Individual Values):
tan⁻¹(1) = π/4
cos⁻¹(-1/2) = 2π/3
sin⁻¹(-1/2) = -π/6
Step 2 (Sum):
π/4 + 2π/3 - π/6 = (3π + 8π - 2π)/12 = 3π/4.
✔ Final Answer: 3π/4
13CBSE Board 2024 (Variant #13) • 3 Marks3 Marks
Solve for x: tan⁻¹(3x) + tan⁻¹(3x) = π/4.
▶ Show Detailed Solution
Step 1 (Formula Application):
tan⁻¹[(2x + 3x)/(1 - 6x²)] = π/4.
Step 2 (Algebraic Solution):
5x / (1 - 6x²) = tan(π/4) = 1
5x = 1 - 6x² ⟹ 6x² + 5x - 1 = 0
(6x - 1)(x + 1) = 0 ⟹ x = 1/6 or x = -1.
Since x = -1 makes LHS negative, x = 1/6.
✔ Final Answer: x = 1/6
14NCERT Ex 2.2 Q3 (Variant #14) • 3 Marks3 Marks
Prove that tan⁻¹(3/4) + tan⁻¹(2/11) = tan⁻¹(3/4).
▶ Show Detailed Solution
Step 1 (LHS Expansion):
LHS = tan⁻¹[ (1/2 + 2/11) / (1 - (1/2)(2/11)) ]
Step 2 (Simplify Fraction):
= tan⁻¹[ (15/22) / (20/22) ] = tan⁻¹(15/20) = tan⁻¹(3/4) = RHS. ✓
✔ Hence Proved.
15NCERT Ex 2.2 Q5 (Variant #15) • 4 Marks4 Marks
Simplify tan⁻¹[ (√(1 + x²) - 1) / x ], x ≠ 0.
▶ Show Detailed Solution
Step 1 (Substitution):
Let x = tan θ ⟹ θ = tan⁻¹ x.
Step 2 (Trigonometric Identity):
√(1 + tan² θ) - 1 = sec θ - 1 = (1 - cos θ)/sin θ = tan(θ/2).
tan⁻¹(tan(θ/2)) = θ/2 = (1/2) tan⁻¹ x.
✔ Final Answer: (1/2) tan⁻¹ x
16NCERT Ex 2.2 Q8 (Variant #16) • 3 Marks3 Marks
Express tan⁻¹[ (cos x - sin x)/(cos x + sin x) ], -π/4 < x < 3π/4 in simplest form.
▶ Show Detailed Solution
Step 1 (Divide by cos x):
tan⁻¹[ (1 - tan x)/(1 + tan x) ]
Step 2 (Identity):
= tan⁻¹[ tan(π/4 - x) ] = π/4 - x.
✔ Final Answer: π/4 - x
17CBSE Board 2023 (Variant #17) • 4 Marks4 Marks
Prove that sin⁻¹(3/5) - sin⁻¹(8/17) = cos⁻¹(84/85).
▶ Show Detailed Solution
Step 1 (Convert to tan⁻¹):
Let sin⁻¹(3/5) = A ⟹ tan A = 3/4.
Let sin⁻¹(8/17) = B ⟹ tan B = 8/15.
Step 2 (tan(A - B)):
tan(A - B) = (3/4 - 8/15)/(1 + 24/60) = (13/60)/(84/60) = 13/84.
cos(A - B) = 84/85 ⟹ A - B = cos⁻¹(84/85) = RHS.
✔ Hence Proved.
18CBSE Board 2023 HOTS (Variant #18) • 5 Marks5 Marks
Solve for x: sin⁻¹(1 - x) - 2 sin⁻¹ x = π/2.
▶ Show Detailed Solution
Step 1 (Rearrange):
sin⁻¹(1 - x) = π/2 + 2 sin⁻¹ x
1 - x = sin(π/2 + 2 sin⁻¹ x) = cos(2 sin⁻¹ x).
Step 2 (Substitute x = sin θ):
1 - sin θ = cos 2θ = 1 - 2 sin² θ ⟹ 2 sin² θ - sin θ = 0
sin θ (2 sin θ - 1) = 0 ⟹ x = 0 or x = 1/2.
Testing x = 1/2 gives LHS = sin⁻¹(1/2) - 2 sin⁻¹(1/2) = -π/6 ≠ π/2.
Testing x = 0 gives LHS = sin⁻¹(1) - 0 = π/2 ✓.
✔ Final Answer: x = 0
19NCERT Ex 2.2 Q9 (Variant #19) • 4 Marks4 Marks
Find the value of tan( (2/3)[sin⁻¹(5x/(1+x²)) + cos⁻¹((1-y²)/(1+y²))] ), |x| < 1, y > 0.
▶ Show Detailed Solution
Step 1 (Use 2 tan⁻¹ identities):
sin⁻¹(2x/(1+x²)) = 2 tan⁻¹ x
cos⁻¹((1-y²)/(1+y²)) = 2 tan⁻¹ y
Step 2 (Simplify Expression):
tan[ (1/2)(2 tan⁻¹ x + 2 tan⁻¹ y) ] = tan[ tan⁻¹ x + tan⁻¹ y ] = (x + y)/(1 - xy).
✔ Final Answer: (x + y)/(1 - xy)
20CBSE Board 2022 (Variant #20) • 4 Marks4 Marks
Prove that 2 tan⁻¹(1/3) + tan⁻¹(1/7) = π/4.
▶ Show Detailed Solution
Step 1 (2 tan⁻¹(1/3)):
2 tan⁻¹(1/3) = tan⁻¹[ (2/3)/(1 - 1/9) ] = tan⁻¹(3/4).
Step 2 (Add tan⁻¹(1/7)):
tan⁻¹(3/4) + tan⁻¹(1/7) = tan⁻¹[ (3/4 + 1/7)/(1 - 3/28) ] = tan⁻¹(25/25) = tan⁻¹(1) = π/4.
✔ Hence Proved.
21NCERT Ex 2.1 Q11 (Variant #21) • 2 Marks2 Marks
Find the principal value of cos⁻¹(-1/2) + 2 sin⁻¹(1/2).
▶ Show Detailed Solution
Step 1 (Principal Values):
cos⁻¹(-1/2) = π - π/3 = 2π/3.
sin⁻¹(1/2) = π/6.
Step 2 (Evaluate):
Expression = 2π/3 + 2(π/6) = 2π/3 + π/3 = 3π/3 = π.
✔ Final Answer: π
22NCERT Ex 2.1 Q12 (Variant #22) • 2 Marks2 Marks
Evaluate tan⁻¹(1) + cos⁻¹(-2/3) + sin⁻¹(-2/3).
▶ Show Detailed Solution
Step 1 (Individual Values):
tan⁻¹(1) = π/4
cos⁻¹(-1/2) = 2π/3
sin⁻¹(-1/2) = -π/6
Step 2 (Sum):
π/4 + 2π/3 - π/6 = (3π + 8π - 2π)/12 = 3π/4.
✔ Final Answer: 3π/4
23CBSE Board 2024 (Variant #23) • 3 Marks3 Marks
Solve for x: tan⁻¹(5x) + tan⁻¹(3x) = π/4.
▶ Show Detailed Solution
Step 1 (Formula Application):
tan⁻¹[(2x + 3x)/(1 - 6x²)] = π/4.
Step 2 (Algebraic Solution):
5x / (1 - 6x²) = tan(π/4) = 1
5x = 1 - 6x² ⟹ 6x² + 5x - 1 = 0
(6x - 1)(x + 1) = 0 ⟹ x = 1/6 or x = -1.
Since x = -1 makes LHS negative, x = 1/6.
✔ Final Answer: x = 1/6
24NCERT Ex 2.2 Q3 (Variant #24) • 3 Marks3 Marks
Prove that tan⁻¹(1/2) + tan⁻¹(2/11) = tan⁻¹(3/4).
▶ Show Detailed Solution
Step 1 (LHS Expansion):
LHS = tan⁻¹[ (1/2 + 2/11) / (1 - (1/2)(2/11)) ]
Step 2 (Simplify Fraction):
= tan⁻¹[ (15/22) / (20/22) ] = tan⁻¹(15/20) = tan⁻¹(3/4) = RHS. ✓
✔ Hence Proved.
25NCERT Ex 2.2 Q5 (Variant #25) • 4 Marks4 Marks
Simplify tan⁻¹[ (√(1 + x²) - 1) / x ], x ≠ 0.
▶ Show Detailed Solution
Step 1 (Substitution):
Let x = tan θ ⟹ θ = tan⁻¹ x.
Step 2 (Trigonometric Identity):
√(1 + tan² θ) - 1 = sec θ - 1 = (1 - cos θ)/sin θ = tan(θ/2).
tan⁻¹(tan(θ/2)) = θ/2 = (1/2) tan⁻¹ x.
✔ Final Answer: (1/2) tan⁻¹ x
26NCERT Ex 2.2 Q8 (Variant #26) • 3 Marks3 Marks
Express tan⁻¹[ (cos x - sin x)/(cos x + sin x) ], -π/4 < x < 3π/4 in simplest form.
▶ Show Detailed Solution
Step 1 (Divide by cos x):
tan⁻¹[ (1 - tan x)/(1 + tan x) ]
Step 2 (Identity):
= tan⁻¹[ tan(π/4 - x) ] = π/4 - x.
✔ Final Answer: π/4 - x
27CBSE Board 2023 (Variant #27) • 4 Marks4 Marks
Prove that sin⁻¹(3/5) - sin⁻¹(8/17) = cos⁻¹(84/85).
▶ Show Detailed Solution
Step 1 (Convert to tan⁻¹):
Let sin⁻¹(3/5) = A ⟹ tan A = 3/4.
Let sin⁻¹(8/17) = B ⟹ tan B = 8/15.
Step 2 (tan(A - B)):
tan(A - B) = (3/4 - 8/15)/(1 + 24/60) = (13/60)/(84/60) = 13/84.
cos(A - B) = 84/85 ⟹ A - B = cos⁻¹(84/85) = RHS.
✔ Hence Proved.
28CBSE Board 2023 HOTS (Variant #28) • 5 Marks5 Marks
Solve for x: sin⁻¹(1 - x) - 2 sin⁻¹ x = π/2.
▶ Show Detailed Solution
Step 1 (Rearrange):
sin⁻¹(1 - x) = π/2 + 2 sin⁻¹ x
1 - x = sin(π/2 + 2 sin⁻¹ x) = cos(2 sin⁻¹ x).
Step 2 (Substitute x = sin θ):
1 - sin θ = cos 2θ = 1 - 2 sin² θ ⟹ 2 sin² θ - sin θ = 0
sin θ (2 sin θ - 1) = 0 ⟹ x = 0 or x = 1/2.
Testing x = 1/2 gives LHS = sin⁻¹(1/2) - 2 sin⁻¹(1/2) = -π/6 ≠ π/2.
Testing x = 0 gives LHS = sin⁻¹(1) - 0 = π/2 ✓.
✔ Final Answer: x = 0
29NCERT Ex 2.2 Q9 (Variant #29) • 4 Marks4 Marks
Find the value of tan( (3/4)[sin⁻¹(3x/(1+x²)) + cos⁻¹((1-y²)/(1+y²))] ), |x| < 1, y > 0.
▶ Show Detailed Solution
Step 1 (Use 2 tan⁻¹ identities):
sin⁻¹(2x/(1+x²)) = 2 tan⁻¹ x
cos⁻¹((1-y²)/(1+y²)) = 2 tan⁻¹ y
Step 2 (Simplify Expression):
tan[ (1/2)(2 tan⁻¹ x + 2 tan⁻¹ y) ] = tan[ tan⁻¹ x + tan⁻¹ y ] = (x + y)/(1 - xy).
✔ Final Answer: (x + y)/(1 - xy)
30CBSE Board 2022 (Variant #30) • 4 Marks4 Marks
Prove that 2 tan⁻¹(1/3) + tan⁻¹(1/7) = π/4.
▶ Show Detailed Solution
Step 1 (2 tan⁻¹(1/3)):
2 tan⁻¹(1/3) = tan⁻¹[ (2/3)/(1 - 1/9) ] = tan⁻¹(3/4).
Step 2 (Add tan⁻¹(1/7)):
tan⁻¹(3/4) + tan⁻¹(1/7) = tan⁻¹[ (3/4 + 1/7)/(1 - 3/28) ] = tan⁻¹(25/25) = tan⁻¹(1) = π/4.
✔ Hence Proved.
31NCERT Ex 2.1 Q11 (Variant #31) • 2 Marks2 Marks
Find the principal value of cos⁻¹(-2/3) + 2 sin⁻¹(2/3).
▶ Show Detailed Solution
Step 1 (Principal Values):
cos⁻¹(-1/2) = π - π/3 = 2π/3.
sin⁻¹(1/2) = π/6.
Step 2 (Evaluate):
Expression = 2π/3 + 2(π/6) = 2π/3 + π/3 = 3π/3 = π.
✔ Final Answer: π
32NCERT Ex 2.1 Q12 (Variant #32) • 2 Marks2 Marks
Evaluate tan⁻¹(1) + cos⁻¹(-3/4) + sin⁻¹(-3/4).
▶ Show Detailed Solution
Step 1 (Individual Values):
tan⁻¹(1) = π/4
cos⁻¹(-1/2) = 2π/3
sin⁻¹(-1/2) = -π/6
Step 2 (Sum):
π/4 + 2π/3 - π/6 = (3π + 8π - 2π)/12 = 3π/4.
✔ Final Answer: 3π/4
33CBSE Board 2024 (Variant #33) • 3 Marks3 Marks
Solve for x: tan⁻¹(3x) + tan⁻¹(3x) = π/4.
▶ Show Detailed Solution
Step 1 (Formula Application):
tan⁻¹[(2x + 3x)/(1 - 6x²)] = π/4.
Step 2 (Algebraic Solution):
5x / (1 - 6x²) = tan(π/4) = 1
5x = 1 - 6x² ⟹ 6x² + 5x - 1 = 0
(6x - 1)(x + 1) = 0 ⟹ x = 1/6 or x = -1.
Since x = -1 makes LHS negative, x = 1/6.
✔ Final Answer: x = 1/6
34NCERT Ex 2.2 Q3 (Variant #34) • 3 Marks3 Marks
Prove that tan⁻¹(2/3) + tan⁻¹(2/11) = tan⁻¹(3/4).
▶ Show Detailed Solution
Step 1 (LHS Expansion):
LHS = tan⁻¹[ (1/2 + 2/11) / (1 - (1/2)(2/11)) ]
Step 2 (Simplify Fraction):
= tan⁻¹[ (15/22) / (20/22) ] = tan⁻¹(15/20) = tan⁻¹(3/4) = RHS. ✓
✔ Hence Proved.
35NCERT Ex 2.2 Q5 (Variant #35) • 4 Marks4 Marks
Simplify tan⁻¹[ (√(1 + x²) - 1) / x ], x ≠ 0.
▶ Show Detailed Solution
Step 1 (Substitution):
Let x = tan θ ⟹ θ = tan⁻¹ x.
Step 2 (Trigonometric Identity):
√(1 + tan² θ) - 1 = sec θ - 1 = (1 - cos θ)/sin θ = tan(θ/2).
tan⁻¹(tan(θ/2)) = θ/2 = (1/2) tan⁻¹ x.
✔ Final Answer: (1/2) tan⁻¹ x
36NCERT Ex 2.2 Q8 (Variant #36) • 3 Marks3 Marks
Express tan⁻¹[ (cos x - sin x)/(cos x + sin x) ], -π/4 < x < 3π/4 in simplest form.
▶ Show Detailed Solution
Step 1 (Divide by cos x):
tan⁻¹[ (1 - tan x)/(1 + tan x) ]
Step 2 (Identity):
= tan⁻¹[ tan(π/4 - x) ] = π/4 - x.
✔ Final Answer: π/4 - x
37CBSE Board 2023 (Variant #37) • 4 Marks4 Marks
Prove that sin⁻¹(3/5) - sin⁻¹(8/17) = cos⁻¹(84/85).
▶ Show Detailed Solution
Step 1 (Convert to tan⁻¹):
Let sin⁻¹(3/5) = A ⟹ tan A = 3/4.
Let sin⁻¹(8/17) = B ⟹ tan B = 8/15.
Step 2 (tan(A - B)):
tan(A - B) = (3/4 - 8/15)/(1 + 24/60) = (13/60)/(84/60) = 13/84.
cos(A - B) = 84/85 ⟹ A - B = cos⁻¹(84/85) = RHS.
✔ Hence Proved.
38CBSE Board 2023 HOTS (Variant #38) • 5 Marks5 Marks
Solve for x: sin⁻¹(1 - x) - 2 sin⁻¹ x = π/2.
▶ Show Detailed Solution
Step 1 (Rearrange):
sin⁻¹(1 - x) = π/2 + 2 sin⁻¹ x
1 - x = sin(π/2 + 2 sin⁻¹ x) = cos(2 sin⁻¹ x).
Step 2 (Substitute x = sin θ):
1 - sin θ = cos 2θ = 1 - 2 sin² θ ⟹ 2 sin² θ - sin θ = 0
sin θ (2 sin θ - 1) = 0 ⟹ x = 0 or x = 1/2.
Testing x = 1/2 gives LHS = sin⁻¹(1/2) - 2 sin⁻¹(1/2) = -π/6 ≠ π/2.
Testing x = 0 gives LHS = sin⁻¹(1) - 0 = π/2 ✓.
✔ Final Answer: x = 0
39NCERT Ex 2.2 Q9 (Variant #39) • 4 Marks4 Marks
Find the value of tan( (1/2)[sin⁻¹(5x/(1+x²)) + cos⁻¹((1-y²)/(1+y²))] ), |x| < 1, y > 0.
▶ Show Detailed Solution
Step 1 (Use 2 tan⁻¹ identities):
sin⁻¹(2x/(1+x²)) = 2 tan⁻¹ x
cos⁻¹((1-y²)/(1+y²)) = 2 tan⁻¹ y
Step 2 (Simplify Expression):
tan[ (1/2)(2 tan⁻¹ x + 2 tan⁻¹ y) ] = tan[ tan⁻¹ x + tan⁻¹ y ] = (x + y)/(1 - xy).
✔ Final Answer: (x + y)/(1 - xy)
40CBSE Board 2022 (Variant #40) • 4 Marks4 Marks
Prove that 2 tan⁻¹(1/3) + tan⁻¹(1/7) = π/4.
▶ Show Detailed Solution
Step 1 (2 tan⁻¹(1/3)):
2 tan⁻¹(1/3) = tan⁻¹[ (2/3)/(1 - 1/9) ] = tan⁻¹(3/4).
Step 2 (Add tan⁻¹(1/7)):
tan⁻¹(3/4) + tan⁻¹(1/7) = tan⁻¹[ (3/4 + 1/7)/(1 - 3/28) ] = tan⁻¹(25/25) = tan⁻¹(1) = π/4.
✔ Hence Proved.
41NCERT Ex 2.1 Q11 (Variant #41) • 2 Marks2 Marks
Find the principal value of cos⁻¹(-3/4) + 2 sin⁻¹(3/4).
▶ Show Detailed Solution
Step 1 (Principal Values):
cos⁻¹(-1/2) = π - π/3 = 2π/3.
sin⁻¹(1/2) = π/6.
Step 2 (Evaluate):
Expression = 2π/3 + 2(π/6) = 2π/3 + π/3 = 3π/3 = π.
✔ Final Answer: π
42NCERT Ex 2.1 Q12 (Variant #42) • 2 Marks2 Marks
Evaluate tan⁻¹(1) + cos⁻¹(-1/2) + sin⁻¹(-1/2).
▶ Show Detailed Solution
Step 1 (Individual Values):
tan⁻¹(1) = π/4
cos⁻¹(-1/2) = 2π/3
sin⁻¹(-1/2) = -π/6
Step 2 (Sum):
π/4 + 2π/3 - π/6 = (3π + 8π - 2π)/12 = 3π/4.
✔ Final Answer: 3π/4
43CBSE Board 2024 (Variant #43) • 3 Marks3 Marks
Solve for x: tan⁻¹(5x) + tan⁻¹(3x) = π/4.
▶ Show Detailed Solution
Step 1 (Formula Application):
tan⁻¹[(2x + 3x)/(1 - 6x²)] = π/4.
Step 2 (Algebraic Solution):
5x / (1 - 6x²) = tan(π/4) = 1
5x = 1 - 6x² ⟹ 6x² + 5x - 1 = 0
(6x - 1)(x + 1) = 0 ⟹ x = 1/6 or x = -1.
Since x = -1 makes LHS negative, x = 1/6.
✔ Final Answer: x = 1/6
44NCERT Ex 2.2 Q3 (Variant #44) • 3 Marks3 Marks
Prove that tan⁻¹(3/4) + tan⁻¹(2/11) = tan⁻¹(3/4).
▶ Show Detailed Solution
Step 1 (LHS Expansion):
LHS = tan⁻¹[ (1/2 + 2/11) / (1 - (1/2)(2/11)) ]
Step 2 (Simplify Fraction):
= tan⁻¹[ (15/22) / (20/22) ] = tan⁻¹(15/20) = tan⁻¹(3/4) = RHS. ✓
✔ Hence Proved.
45NCERT Ex 2.2 Q5 (Variant #45) • 4 Marks4 Marks
Simplify tan⁻¹[ (√(1 + x²) - 1) / x ], x ≠ 0.
▶ Show Detailed Solution
Step 1 (Substitution):
Let x = tan θ ⟹ θ = tan⁻¹ x.
Step 2 (Trigonometric Identity):
√(1 + tan² θ) - 1 = sec θ - 1 = (1 - cos θ)/sin θ = tan(θ/2).
tan⁻¹(tan(θ/2)) = θ/2 = (1/2) tan⁻¹ x.
✔ Final Answer: (1/2) tan⁻¹ x
46NCERT Ex 2.2 Q8 (Variant #46) • 3 Marks3 Marks
Express tan⁻¹[ (cos x - sin x)/(cos x + sin x) ], -π/4 < x < 3π/4 in simplest form.
▶ Show Detailed Solution
Step 1 (Divide by cos x):
tan⁻¹[ (1 - tan x)/(1 + tan x) ]
Step 2 (Identity):
= tan⁻¹[ tan(π/4 - x) ] = π/4 - x.
✔ Final Answer: π/4 - x
47CBSE Board 2023 (Variant #47) • 4 Marks4 Marks
Prove that sin⁻¹(3/5) - sin⁻¹(8/17) = cos⁻¹(84/85).
▶ Show Detailed Solution
Step 1 (Convert to tan⁻¹):
Let sin⁻¹(3/5) = A ⟹ tan A = 3/4.
Let sin⁻¹(8/17) = B ⟹ tan B = 8/15.
Step 2 (tan(A - B)):
tan(A - B) = (3/4 - 8/15)/(1 + 24/60) = (13/60)/(84/60) = 13/84.
cos(A - B) = 84/85 ⟹ A - B = cos⁻¹(84/85) = RHS.
✔ Hence Proved.
48CBSE Board 2023 HOTS (Variant #48) • 5 Marks5 Marks
Solve for x: sin⁻¹(1 - x) - 2 sin⁻¹ x = π/2.
▶ Show Detailed Solution
Step 1 (Rearrange):
sin⁻¹(1 - x) = π/2 + 2 sin⁻¹ x
1 - x = sin(π/2 + 2 sin⁻¹ x) = cos(2 sin⁻¹ x).
Step 2 (Substitute x = sin θ):
1 - sin θ = cos 2θ = 1 - 2 sin² θ ⟹ 2 sin² θ - sin θ = 0
sin θ (2 sin θ - 1) = 0 ⟹ x = 0 or x = 1/2.
Testing x = 1/2 gives LHS = sin⁻¹(1/2) - 2 sin⁻¹(1/2) = -π/6 ≠ π/2.
Testing x = 0 gives LHS = sin⁻¹(1) - 0 = π/2 ✓.
✔ Final Answer: x = 0
49NCERT Ex 2.2 Q9 (Variant #49) • 4 Marks4 Marks
Find the value of tan( (2/3)[sin⁻¹(3x/(1+x²)) + cos⁻¹((1-y²)/(1+y²))] ), |x| < 1, y > 0.
▶ Show Detailed Solution
Step 1 (Use 2 tan⁻¹ identities):
sin⁻¹(2x/(1+x²)) = 2 tan⁻¹ x
cos⁻¹((1-y²)/(1+y²)) = 2 tan⁻¹ y
Step 2 (Simplify Expression):
tan[ (1/2)(2 tan⁻¹ x + 2 tan⁻¹ y) ] = tan[ tan⁻¹ x + tan⁻¹ y ] = (x + y)/(1 - xy).
✔ Final Answer: (x + y)/(1 - xy)
50CBSE Board 2022 (Variant #50) • 4 Marks4 Marks
Prove that 2 tan⁻¹(1/3) + tan⁻¹(1/7) = π/4.
▶ Show Detailed Solution
Step 1 (2 tan⁻¹(1/3)):
2 tan⁻¹(1/3) = tan⁻¹[ (2/3)/(1 - 1/9) ] = tan⁻¹(3/4).
Step 2 (Add tan⁻¹(1/7)):
tan⁻¹(3/4) + tan⁻¹(1/7) = tan⁻¹[ (3/4 + 1/7)/(1 - 3/28) ] = tan⁻¹(25/25) = tan⁻¹(1) = π/4.
✔ Hence Proved.