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CBSE CLASS X Mathematics Trignometry Two mark questions with answers Q1. Prove the identity cos/(1 + sin) = (1 - sin)/cos Ans1. L.H.S. = cos/(1 + sin) = {cos/(1 + sin)} x {(1 - sin)/(1 - sin)} = {cos(1 - sin)}/(1 - sin2) = cos(1 - sin)/cos2 = (1 - sin)/cos = R.H.S. Q2. Prove that cot + tan = cosec.sec Ans2. L.H.S. = cot + tan = (cos/sin) + (sin/cos) = (cos2 + sin2)/(sin.cos) = 1/(sin.cos).......[... cos2 + sin2 = 1] = cosec.sec = R.H.S. Q3. Prove that (cos2/sin) + sin = cosec Ans3. L.H.S. = (cos2/sin) + sin = (cos2 + sin2)/sin.................[... cos2 + sin2 = 1] = 1/sin = cosec = R.H.S. Q4. Prove the identity {1/(1 + sinA)} + {1/(1 - sinA)} = 2sec2A Ans4. {1/(1 + sinA)} + {1/(1 - sinA)} = (1 - sinA + 1 + sinA)/(1 - sin2A) = 2/cos2A..........[... cos2A + sin2A = 1] = 2sec2A = R.H.S. Q5. Prove that (sec - 1)/(sec + 1) = (1 - cos)/(1 + cos) Ans5. L.H.S. = (sec - 1)/(sec + 1) = {(1/cos) - 1}/{(1/cos) + 1} = {(1 - cos)/cos}/{(1 + cos)/cos} = (1 - cos)/(1 + cos) = R.H.S. Q6. Prove that (1 + cos)/sin2 = 1/(1 - cos). Ans6. L.H.S. = (1 + cos)/sin2 = {(1 + cos)/sin2} x {(1 - cos)/(1 - cos)} = (1 - cos2)/{sin2(1 - cos)} = sin2/{sin2(1 - cos)}...................[... cos2 + sin2 = 1] = 1/(1 - cos) = R.H.S. Q7. Prove the identity cos2 + cos2.cot2 = cot2 Ans7. L.H.S. = cos2 + cos2.cot2 = cos2(1 + cot2) = cosec2cos2...............[... 1 + cot2 = cosec2] = cos2(1/sin2)= cot2 = R.H.S. Q8. Prove that sin2(1 + cot2) = 1 Ans8. L.H.S. = sin2(1 + cot2) = sin2.cosec2........................[... 1 + cot2 = cosec2] = sin2(1/sin2) = 1 = R.H.S. Q9. Simplify cos2A + {1/(1 + cot2A)}. Ans9. cos2A + {1/(1 + cot2A)} = cos2A + (1/cosec2A)...........[... 1 + cot2A = cosec2A] = cos2A + sin2A =1 Hence cos2A + {1/(1 + cot2A)} = 1 Q10. Prove that sec2 + cosec2 = sec2cosec2. Ans10. L.H.S. = sec2 + cosec2 = (1/cos2) + (1/sin2) = (sin2 + cos2)/(sin2cos2) = 1/cos2sin2 = sec2cosec2 = R.H.S. Q11.Simplify (1 + tan2A)(1 - sinA)(1 + sinA). Ans11. (1 + tan2A)(1 - sinA)(1 + sinA) = (sec2A)(1 - sin2A)............[... 1 + tan2A = sec2A and sin2A + cos2A = 1] = sec2A.cos2A = (1/cos2A)cos2A = 1 Hence (1 + tan2A)(1 - sinA)(1 + sinA) = 1 Q12. Prove that cot - tan = (2cos2 - 1)/(sin.cos). Ans12. L.H.S. = cot - tan = (cos/sin) - (sin/cos) = (cos2 - sin2)/(sin.cos) = [cos2 - (1 - cos2)]/(sin.cos).......[... sin2 + cos2 = 1] = (2cos2 - 1)/(sin.cos) = R.H.S. Q13. Prove that tan2 - sin2 = tan2sin2. Ans13. L.H.S. = tan2 - sin2 = (sin2/cos2) - sin2 = sin2[(1/cos2) - 1] = sin2{(1 - cos2)/cos2}.........[... sin2 + cos2 = 1] = (sin2.sin2)/cos2 = tan2.sin2 = R.H.S. Q14. Prove that (sin2A.cot2A) + (cos2A.tan2A) = 1. Ans14. L.H.S. = sin2A.cot2A + cos2A.tan2A = sin2A.(cos2A/sin2A) + cos2A.(sin2A/cos2A) = cos2A + sin2A = 1 = R.H.S. Q15. Prove that identity sinA/(1 - cosA) = (1 + cosA)/sinA. Ans15. L.H.S. = sinA/(1 - cosA) = {sinA/(1 - cosA)} x {(1 + cosA)/(1 + cosA)} = {sinA(1 + cosA)/(1 - cos2A)}..........[... sin2A + cos2A = 1] = sinA(1 + cosA)/sin2A = (1 + cosA)/sinA = R.H.S. Four mark questions with answers Q1. Prove that Ans1. = = =0 = R.H.S Q2. Prove that Ans2. = cosA + sinA = R.H.S Q3. Prove that Ans3. .........[... sin2 + cos2 = 1] = (sec + 1)/(sec - 1) = R.H.S Q4. Prove that Ans4. ............[... 1 + tan2 = sec2, 1 + cot2 = cosec2] = = = = = (sec2 .cosec2) (1/cos2).(1/sin2) 1/(1 - sin2)sin2..........[... sin2 + cos2 = 1] 1/(sin2 - sin4 R.H.S Q5. Prove that (sin + cosec)2 + (cos + sec)2 = 7 + tan2 + cot2. Ans5. L.H.S = (sin + cosec)2 + (cos + sec)2 sin2 + cosec2 + 2sin.cosec + cos2 + sec2 + 2cos.sec = (sin2 + cos2) + cosec2 + sec2 + 2sin.(1/sin) + 2cos.(1/cos) = 1 + cosec2 + sec2 + 2 + 2...........[... sin2 + cos2 = 1] = 5 + 1 + cot2 + 1 + tan2 = 7 + cot2 + tan2 = R.H.S. Q6. Prove that Ans6. = [(sin3A - cos3A)/{sinA.cosA(sinA - cosA)}] = [(sinA - cosA)(sin2A + cos2A + sinA.cosA)/(sinA - cosA)sinA cosA]............[... a3 b3 = (a - b)(a2 + b2 + ab)] = (sin2A + cos2A + sinA.cosA)/(sinA.cosA) = (1 + sinA.cosA)/(sinA.cosA)...........[... sin2A + cos2A = 1] = (1/sinA.cosA) + 1 = secA.cosecA + 1 = R.H.S Q7. Prove that 2(sin6 + cos6) - 3(sin4 + cos4) + 1 = 0 Ans7. L.H.S = 2(sin6 + cos6) - 3(sin4 + cos4) + = 2[(sin2 + cos2)3 - 3sin2.cos2(sin2 + cos2) - 3[(sin2 + cos2)2 - 2sin2.cos2] + 1.......[... a3 + b3 = (a + b)3 - 3ab(a + b) & a2 + b2 = (a + b)2 - 2ab] = 2[12 - 3sin2.cos2.1] - 3[12 - 2sin2.cosec2 + 1)......[... sin2 + cos2 = 1] = 2 - 6sin2.cos2 - 3 + 6sin2.cos2 + 1 = 2 - 3 + 1 = 0 = R.H.S Q8. Show that Ans8. L.H.S = [{(1)/(cosec - cot)} - {(1/sin)}] = = [{(sin)/(1 - cos)} - (1/sin)] = [(sin2 - 1 + cos)/{sin(1 - cos)}] = [cos - (1 - sin2)]/[sin(1 - cos)] = {cos - cos2}/{sin(1 - cos)}....[... sin2 + cos2 = 1] = {cos(1 - cos)}/{sin(1 - cos)} = cos/sin = cot ..............(i) R.H.S. = (1/sin) - {1/(cosec + cot)} = (1/sin) - [1/{(1/sin) + (cos/sin)}] = (1/sin) - {sin/(1 + cos)} = (1 + cos - sin2)/{sin(1 + cos)} = (cos2 + cos)/{sin(1 + cos)}.......[... sin2 + cos2 = 1] = {cos(1 + cos)}/{sin(1 + cos)} = cos/sin = cot ................(ii) From (i) and (ii) L.H.S. = R.H.S. Q9. Prove that (tanA + secA - 1)/(tanA - secA + 1) = (1 + sinA)/cosA. Ans9. L.H.S. = (tanA + secA - 1)/(tanA - secA + 1) = {tanA + secA - (sec2A - tan2A)}/(tanA - secA + 1).....[Putting 1 = sec2A - tan2A as 1 + tan2A = sec2A] = {tanA + secA - (secA + tanA)(secA - tanA)}/(tanA - secA + 1) = {(secA + tanA) (1 - secA + tanA)}/(tanA - secA + 1) = secA + tanA = (1/cosA) + sinA/cosA) = (1 + sinA)/cosA = R.H.S. Q10. Prove that + = 2cosec. Ans10. L.H.S. = = + + = + = [(1 + cos)/sin] + [(1 - cos)/sin].....[... sin2 + cos2 = 1] = (1 + cos + 1 - cos)/sin = 2/sin = 2cosec = R.H.S. Hence the required result. Q11. Prove that tan - cot = (2sin2 - 1)/(sin.cos) Ans11. L.H.S. = tan - cot = (sin/cos) - (cos/sin) = (sin2 - cos2)/(sincos) = (sin2 - 1 + sin2)/(sin.cos)...........[... sin2 + cos2 = 1] = (2sin2 - 1)/(sin.cos) = R.H.S. Q12. Prove that sec4 - sec2 = tan4 + tan2. Ans12. L.H.S. sec4 - sec2 = sec2[sec2 - 1] = (1 + tan2)[tan2].............[... sec2 - tan2 = 1] = tan2 + tan4 = R.H.S. Six mark questions with answers Q1. If cot = n cot and cos = mcos, then prove that sin2 = (m2 - 1)/(n2 - 1). Ans1. cot = n cot cot = cot/n tan = n/cot..............(1) cos = mcos cos = cos/m sec = m/cos..............(2) Squaring (1) and (2) and subtracting, we get tan2 - sec2 = (n2/cot2) - (m2/cos2) -1 = (n2sin2/cos2) - (m2/cos2)......[... sec2 = 1 + tan2] (m2/cos2) - (n2sin2/cos2) = 1 (1/cos2)[m2 - n2sin2] = 1 m2 - n2sin2 = cos2 m2 - n2sin2 = 1 - sin2.....[... sin2 + cos2 = 1] m2 - 1 = n2sin2 - sin2 m2 - 1 = (n2 - 1)sin2 sin2 = (m2 - 1)/(n2 - 1) Hence Proved. Q2. The angle of elevation of a tower at a point is 45o. After going 40 m towards the foot of the tower, the angle of elevation of the tower becomes 60o. Find the height of the tower. Ans2. Let AB = h metres be the height of the tower. Let C & D be the points where the angles subtended are 450 and 60o. CD = 40 m Let DB = x m and BC = CD + DB = (40 + x)m In ABD tan 60o = AB/BD 3 = h/x h = 3x x = h/3.................(1) In ABC tan45o = AB/BC 1 = h/(40 + x) h = 40 + x Putting the value of x from (1), we get h = 40 + h/3 h - (h/3) = 40 h(1 - 1/3) = 40 h{(3 - 1)/3} = 40 h = 403/(3 - 1) = 403/(3 - 1) x {(3 + 1)/(3 + 1)} = 40(3 + 3)/(3 - 1) = 40(3 + 3)/2 = 20(3 + 1.732) = 20 x 4.732 = 94.64 m Hence, the height of the tower is 94.64 m. Q3. From the top of a building 60 m high, the angles of depression of the top and bottom of a tower are observed to be 30o and 60o. Find the height of the tower. Ans3. Let h be the height of the tower AB = 60 m Suppose BD = t m In ABD tan 60o = AB/BD 3 = 60/t t = 60/3................(1) In AEC tan 30o = AE/CE tan 30o = AE/BD..................[... CE = BD] 1/3 = AE/t 1/3 = AE/(60/3)........[Putting the value of t from (1)] AE = (1/3) x (60/3) AE = 20 m Now h = CD = EB = AB - AE = (60 - 20)m = 40 m Thus, the height of the tower = 40 m. Q4. The horizontal distance between two towers is 80 m. The angle of depression of the top of first tower when seen from the top of the second tower is 35 o. If the height of the second tower is 160 m, find the height of the first tower. Ans4. Let AB be the tower 160 m high. Let h be the height of the other tower DE Now BD = 80 m EAC = 35o AEF = 35o In AEF tan35o = AF/EF 0.7002 = (AB - FB)/EF 0.7002 = (160 - h)/80 0.7002 x 80 = 160 - h 56.016 = 160 - h h = 160 - 56.016 h = 103.984 Height of the tower = 103.984 m.