Trigonometry Practice Sheet Warrior 2025 PDF

Warrior 2025 Practice Sheet TRIGONOMETRY 1. If the value of cos A = 4/5, then tan A = ? tan x 8....

Warrior 2025 Practice Sheet TRIGONOMETRY 1. If the value of cos A = 4/5, then tan A = ? tan x 8. =1 (A) 1 (B) 3/4 sin x3 + sin x cos x (C) 4/3 (D) 4/5 cos x (A) true 4sin A − cosA (B) false 2. If 4 tan A = 3, then =? 4sin A + cosA (C) can't say 2 1 (D) None of these (A) (B) 3 3 1 tan 2  (C) (D) 1 9. + 1equals to : 2 1 + sec  1 (A) tan θ (B) 1 cos 3. If sin  = , then the value of 2cot2 + 2is: 3 sec  –1 (C) (D) sec θ + tan θ (A) 6 (B) 1 cot  (C) 18 (D) 10 10. If x = a sec cos , y = bsec sin  and z = c tan, 4. (sin 30° + cos 60°) – (sin 60° + cos 30°) is equal to: x 2 y2 (A) 0 (B) 1 then + = a 2 b2 (C) 1– 3 (D) 1 + 3 z2 z2 (A) 2 + a2 (B) 1 – c c2 a 5. If cos x = , then tan x is equal to : z2 z2 b (C) –1 (D) 1 + c2 c2 a2 (A) b2 – b 11. In a ABC, right angled at B, AB = 15 cm, BC = 8 b–a cm. Determine (B) b (i) sin A, cos A b2 – a 2 (ii) sin C, cos C (C) a 1 (b – a) 12. In a ABC, right angled at B, if tan A = , find (D) 3 b the value of (i) sinA cosC + cosA sinC 1 6. In cot  = , the value of sec2 + cosec2 is (ii) cosA cosC – sinA sinC. 3 40 (A) 1 (B) 1– tan 2 A 9 13. If 3 cotA = 4, check whether 1 + tan 2 A 38 1 (C) (D) 5 = cos2 A – sin2 A or not. 9 3 1 + tan  14. Find the value of each of the following: 7. Given that sec  = 2 , the value of is: sin  5sin 2 30 + cos2 45 − 4tan 2 30 (i) (A) 2 2 (B) 2 2sin30 cos30 + tan 45 (ii) sin 45° (1 + cot2 45°) + cos2 45°(1 + tan2 45°) 2 (C) 3 2 (D) 2 + sin2 20° + cos2 20° 15. Find the value of  in each of the following : 18. Prove that: tan  cot  (i) 2 sin 2 = 3 + = 1 + sec  cosec  1 − cot  1 − tan  (ii) 2 cos 3 = 1 19. Prove that: tan 2  1– cos  cosA − sin A + 1 = = cosecA + cot A 16. Prove that : sec  + 1 cos  cosA + sin A − 1 20. Prove that: 17. Prove that cosec (1 – cos ) + sec (1 – sin ) = 2 2 2 2 2 1 (cosecA − sin A)(secA − cosA) = tan A + cot A  ANSWER KEY 1. (B) 11. (Check Solution) 2. (C) 12. (Check Solution) 3. (C) 13. (Check Solution) 4. (C) 14. (Check Solution) 5. (C) 15. (Check Solution) 6. (D) 16. (Check Solution) 7. (A) 17. (Check Solution) 8. (A) 18. (Check Solution) 9. (B) 19. (Check Solution) 10. (D) 20. (Check Solution)  HINTS AND SOLUTIONS 1. (B) 3/4 12. We know that 1 2. (C) 2 3. (C) 18 BC 1 4. (C) 1– 3 tanA = = AB 3 BC 1 b2 – a 2 tan A = = 5. (C) AB 3 a  BC : AB = 1 : 3 1 Let BC = k and AB = 3k 6. (D) 5 3 Then, AC = AB + BC2 (Pythagoras theorem) 2 7. (A) 2 2 = ( 3k)2 + (k)2 = 3k2 + k2 = 4k 2 = 2k 8. (A) true BC k 1 Now, sin A = = = AC 2k 2 1 13. 3 cot A = 4 [Given] 9. (B) cos 4  cot A = 3 z2 Draw a right triangle ABC, right angled at B. 10. (D) 1 + c2 11. In right ABC, we have AC2 = AB2 + BC2 [By Pythagoras theorem]  AC2 = (15 cm)2 + (8 cm)2 = 289 cm2  AC = 17 cm AB 4 Then, cot A = = BC 3  If AB = 4 k, then BC = 3k, where k is a positive number. In right ABC, we have AC2 = AB2 + BC2 [By Pythagoras theorem] With reference to A, we have  AC2 = (4k)2 + (3k)2 = 16k2 + 9k2 = 25k2 Base AB = 15 cm, perpendicular BC = 8 cm and  AC = 5k BC 3k 3 hypotenuse AC = 17 cm. Now, sinA = = = BC 8 cm 8 AC 5k 5  sin A = = = AB 4k 4 AC 17 cm 17 cosA = = = AB 15 cm 15 AC 5k 5 and cos A = = = BC 3k 3 AC 17 cm 17 and tanA = = = AB 4k 4 With reference to C, we have 2  3 9 Base BC = 8 cm, perpendicular AB = 15 cm and 1 −  4  1− 1 − tan 2 A   hypotenuse AC = 17 cm. LHS = = = 16 1 + tan 2 A   2 9 1 +   1 + 16 AB 15 cm 15 3  sin C = = = AC 17 cm 17 4 BC 8 cm 8 7 16 7 and cos C = = = =  = AC 17 cm 17 16 25 25  5sin 2 30 + cos2 45 − 4tan 2 30 17. LHS = cosec2 (1 – cos2 ) + sec2 (1 – sin2 ) 14. (i) 2sin30 cos30 + tan 45 = cosec2 sin2 + sec2 cos2 [ 1 – cos2 = sin2 2 1  1  2  1  2 5 1 4 and 1 – sin2 = cos2] 5  +   − 4  + − 1 1    2 2  3 = 4 2 3 = cosec2 + sec2 = cosec  2 sec2  1 3 3 2  +1 +1 = 1 + 1 = 2 = RHS 2 2 2 15 + 6 − 16 5 tan  cot  18. + = 1 + sec  cosec  = 12 = 12 1 − cot  1 − tan  3+2 2+ 3 sin  cos  tan  cot  = cos  + sin  2 2 = + 5 2 5 1 − cot  1 − tan  1 − cos  1 − sin  =  = sin  cos  12 2 + 3 6(2 + 3) sin  cos  (ii) sin2 45° (1 + cot2 45°) + cos2 45 (1 + tan2 45°) = cos  + sin  + sin2 20° + cos2 20° sin  − cos  cos  − sin   1   2  1   2 sin  cos  =  1 + (1)  +   1 + (1)  2 2  2  2 sin 2  cos2  = + cos (sin  − cos ) sin (sin  − cos ) + sin2 20 + cos2 20 1  sin 2  cos 2   1 1 =  −  =  2 +  2 + 1 [ sin2 20° + cos2 20° = 1] (sin  − cos )  cos  sin   2 2    sin  − cos   3 3 1 =1+1+1=3 =     sin  − cos    sin  cos   15. (i) we have,  = 1  (   (sin  − cos ) sin  + cos  + sin  cos  2 2 )   2 sin 2 = 3  sin  − cos    sin  cos     3 (1 + sin  cos )  sin 2 = = = sec cosec + 1 = R.H.S. 2 sin  cos   sin 2 = sin 60° 19. L.H.S.  2 = 60°   = 30° cos A sin A − + 1 cot A − 1 + cosecA (ii) we have, = sin A sin A sin A = cos A sin A + + 1 cot A + 1 − cosecA 2 cos 3 = 1 sin A sin A sin A 1 {(cot A) − (1 − cosecA)}{(cot A) − (1 − cosecA)}  cos 3 = = 2 {(cot A) + (1 − cosecA)}{(cot A) − (1 − cosecA)}  cos 3 = cos 60° (cot A − 1 + cosecA)2  3 = 60°   = 20 = (cot A)2 − (1 − cosecA)2 cot 2 A+1+cosec2 A–2cot A–2cosecA+2cot AcosecA = ( cot 2 A– 1+cosec2 A–2cosecA ) tan 2  sec2  − 1 16. LHS = = [ tan2  = sec2  – 1] 2cosec2 A+2cot AcosecA–2cot A–2cosecA sec  + 1 sec  + 1 = (sec  − 1)(sec  + 1) ( cot 2 A– 1+cosec2 A–2cosecA ) = (sec  + 1) (cosecA + cot A)(2cosecA − 2) = −1 − 1 + 2cosecA 1 1 − cos  = sec  – 1 + –1= = RHS (cosecA + cot A)(2cosecA − 2) cos cos  = (2cosecA − 2) = cosec A + cot A = R.H.S. 20. L.H.S. R.H.S. = (cosec A – sin A)(sec A – cos A) 1 1 = =  1  1  tan A + cot A sin A + cos A = − sin A  − cos A   sin A  cos A  cos A sin A  1 − sin 2 A  1 − cos 2 A  1 sin AcosA = 2 = 2 = sin AcosA =   sin A + cos A sin A + cos2 A 2  sin A  cos A  sin AcosA = ( cos A)(sin A) = sin Acos A 2 2 Hence, L.H.S. = R.H.S. sin Acos A PW Web/App - https://smart.link/7wwosivoicgd4 Library- https://smart.link/sdfez8ejd80if