If ? = 3 sin-1 (6 / 11) and ? = 3 cos-1 (4 / 9), where the inverse trigonometric functions take only the principal values, then the correct option (s) is (are) 1) cosβ > 0 2) sinβ < 0 3) cos(α + β) > 0 4) cosα < 0 Solution: (3) cos(α + β) > 0 ɑ = 3 sin-1 (6 / 11) > 3... View Article
The value of cot-1 (9) + cosec-1 (?41 / 4) is 1) π/2 2) π/4 3) π/3 4) π Solution: (2) π / 4 cot-1 (9) + cosec-1 (√41 / 4) = tan-1 (1 / 9) + tan-1 (4 / 5) = tan-1 (1 / 9) + (4... View Article
The value of tan {cos-1 (- 2 / 7) – (Ï€ / 2)} is 1) 2/3√5 2) 2/3 3) 1/√5 4) 4/√5 Solution: (1) 2 / 3√5 tan {cos-1 (- 2 / 7) - (Ï€ / 2)} = tan {Ï€ - cos-1 (2 / 7) - (π / 2)} = tan {(π / 2) -... View Article
If sin-1 (1 – x) – 2sin-1 x = ? / 2, then x is equal to 1) 0, - (1/2) 2) 0, 1/2 3) 0 4) None of these Solution: (3) 0 sin-1 (1 - x) - 2sin-1 x = Ï€ / 2 sin-1 (1 - x) = Ï€ / 2 + 2 sin-1... View Article
The value of x, where x > 0 and tan {sec-1 (1 / x)} = sin (tan-1 2) is 1) √5 2) √5/3 3) 1 4) 2/3 Solution: (2) √5 / 3 Let {sec-1 (1 / x)} = θ sec θ = 1 / x cos θ = x tan {sec-1 (1 / x)} = tan θ = √1... View Article
If 2 tan-1 (cos x) = tan-1 (2 cosec x), then the value of x is 1) 3π/4 2) π/4 3) π/3 4) None of these Solution: (2) π/4 2 tan-1 (cos x) = tan-1 (2 cosec x) tan-1 (2cosx / [1 - cos2 x]) = tan-1... View Article
If sin-1 x = ? / 5, for some x ? (-1, 1), then the value of cos-1 x is 1) 3π/10 2) 5π/10 3) 7π/10 4) 9π/10 Solution: (1) 3π / 10 sin-1 x = π / 5 sin-1 x + cos-1 x = π / 2 (π / 5) + cos-1 x = π / 2... View Article
If sin-1 x + sin-1 y = ? / 2, then cos-1 x + cos-1 y = 1) π/2 2) π/4 3) π 4) 3π/4 Solution: (1) π / 2 sin-1 x + sin-1 y = π / 2 (π / 2) - cos-1 x + (π / 2) - cos-1 y = π / 2 cos-1 x +... View Article
If sin-1 2a / [1 + a2] – cos-1 [1 – b2] / [1 + b2] = tan-1 2x / [1 + x2], then the value of x is 1) a 2) b 3) [a + b] / [1 - ab] 4) [a - b] / [1 + ab] Solution: (4) [a - b] / [1 + ab] sin-1 2a / [1 + a2] - cos-1 [1 - b2] / [1 +... View Article
If ?3 + i = (a + ib) (c + id), then tan-1 (b / a) + tan-1 (d / c) has the value 1) (π / 3) + 2nπ, n ∈ I 2) nπ + (π / 6), n ∈ I 3) nπ - (π / 3), n ∈ I 4) 2nπ - (π / 3), n ∈ I Solution: (2) nπ + (π / 6), n ∈ I √3... View Article
If sin-1 (x / 5) + cosec-1 (5 / 4) = ? / 2, then the value of x is 1) 1 2) 3 3) 4 4) 5 Solution: (2) 3 sin-1 (x / 5) + cosec-1 (5 / 4) = π / 2 sin-1 (x / 5) + sin-1 (4 / 5) = π / 2 sin-1 (x / 5)... View Article
tan-1 x / (?a2 – x2) = 1) 2 sin-1 (x / a) 2) sin-1 (2x / a) 3) sin-1 (x / a) 4) cos-1 (x / a) Solution: (3) sin-1 (x / a) Given, tan-1 x / (√a2 - x2) = θ... View Article
The solution of tan-1 2? + tan-1 3? = ? / 4 is 1) 1/√3 2) 1/3 3) 1/6 4) 1/√6 Solution: (3) ⅙ tan-1 2θ + tan-1 3θ = π / 4 tan-1 (2θ + 3θ) / (1 - 2θ × 3θ) = π / 4 5θ / [1 - 6θ2]... View Article
If sec-1 (?1 + x2) + cosec-1 (?1 + y2) / y + cot-1 (1 / z) = 3?, then x + y + z = 1) xyz 2) 2xyz 3) xyz2 4) x2yz Solution: (1) xyz sec-1 (√1 + x2) + cosec-1 (√1 + y2) / y + cot-1 (1 / z) = 3π cos-1 1 / (√1 + x2)... View Article
The cos-1 x = ?, (0 < x < 1) and sin-1 (2x ?1 – x2) + sec-1 1 / (2x2 – 1) = 2? / 3, then tan-1 (2x) = 1) Ï€/6 2) Ï€/4 3) Ï€/3 4) Ï€/2 Solution: (3) Ï€ / 3 cos-1 x = É‘, (0 < x < 1) x = cos É‘ sin-1 (2x √1 - x2) + sec-1 1 / (2x2 -... View Article
The value of cot (?n=123 cot-1 (1 + ?k=1n 2k)) is 1) 23 / 25 2) 25 / 23 3) 23 / 24 4) 24 / 23 Solution: (2) 25 / 23 ∑k=1n 2k = 2 [1 + 2 + 3 + …. + n] = 2 [(n (n + 1)] / 2 = (n2 +... View Article
The principal value of sin-1 {tan (- 5? / 4)} is 1) π / 4 2) - (π / 4) 3) π / 2 4) - (π / 2) Solution: (4) - (π / 2) sin-1 {tan (- 5π / 4)} = sin-1{- tan (π + π / 4)} = sin-1 (-... View Article
If (tan-1 x)2 + (cot-1 x)2 = 5?2 / 8, then x is equal to 1) 0 2) 2 3) 1 4) - 1 5) 2√2 Solution: (4) - 1 (tan-1 x)2 + (cot-1 x)2 = 5π2 / 8 (tan-1 x)2 + [(π / 2) - tan-1 x]2 = 5π2 / 8... View Article
If 5 cos-1 [1 – x2] / [1 + x2] + 7 sin-1 2x / [1 + x2] – 4 tan-1 2x / [1 – x2] – tan-1 x = 5?, then x = 1) 3 2) -√3 3) √2 4) 2 5) √3 Solution: (5) √3 5 cos-1 [1 - x2] / [1 + x2] + 7 sin-1 2x / [1 + x2] - 4 tan-1 2x / [1 - x2] -... View Article
sin (2 sin-1 ?63 / 65) = 1) 2√126 / 65 2) 4√65 / 65 3) 8√63 / 65 4) √63 / 65 Solution: (1) 2√126 / 65 sin (2 sin-1 √63 / 65) = sin (sin-1 2 * √63 / 65 * √1... View Article
