ID : in-10-trigonometry [1] Class 10 Trigonometry For more such worksheets visit www.edugain.com Answer t he quest ions (1) An equilateral triangle width side of length 18 3 cm is inscribed in a circle. Find the radius of circle. (2) Simplif y (3) [cos(20 ) cos(21 ) cos(22 )... cos(70 ) ] [cosec(20 ) cosec(21 ) cosec(22 ) ] =? (4) Simplif y cos 6 θ + sin 6 θ + 3 cos 2 θ sin 2 θ (5) If AB =, f ind value of BC. (6) Simplif y (1 + tanθ + secθ) (1 + cotθ - cosecθ) Choose correct answer(s) f rom given choice (7) Simplif y sin 6 θ + cos 6 θ + 3sin 2 θ cos 2 θ a. 2 b. 1 c. -1 d. 3 (8) (cotθ + 5) (5 cotθ + 1) =? a. 26 cotθ - cosec 2 θ b. 26 cotθ - 5 cosec 2 θ c. 26 cotθ + 5 cosec 2 θ d. 26 cotθ + cosec 2 θ
(9) cos 2 76 + cos 2 14 sin 2 76 + sin 2 14 + cos 2 14 + cos 76 sin 14 =? ID : in-10-trigonometry [2] a. 2 b. 1 c. 0 d. 3 (10) Simplif y. a. 2cot θ b. 2cosec θ c. 2sec θ d. 2tan θ (11) =? a. sinθ cosθ b. 2 c. sinθ - cosθ d. sinθ + cosθ (12) If tan θ + cot θ = 4, f ind the value of tan 2 θ + cot 2 θ. a. 18 b. 16 c. 14 d. 12 Fill in the blanks (13) If 2sinθ cosθ = 2, the value of sinθ + 2cosθ =. Check True/False (14) The value of tanθ (θ < 90 ) decreases as θ increases. True False (15) tan 87 - cot 87 > 0 True False 2016 Edugain (www.edugain.com). All Rights Reserved Many more such worksheets can be generated at www.edugain.com
Answers ID : in-10-trigonometry [3] (1) 18 cm Triangle ABC is inscribed in a circle. Lets O be the center of circle. Now connect all vertices of the triangle to O, and draw a perpendicular f rom O to side BC of triangle at point D. OB and OC are bisectors of B and C respectively, hence OBD = 30 Since ODB is right angle BD/OB = cos 30 = 3/2 Step 5 OB = BD / ( 3/2) = (9 3) / ( 3/2) = 18 cm (2) 0 On combining two f ractions = = = Since sin 2 θ + cos 2 θ + = 1 =
(3) 1 ID : in-10-trigonometry [4] We need to f ind some of f ollowing series S = (cos(20 ) cos(21 ) cos(22 )... cos(70 ) )(cosec(20 ) cosec(21 ) cosec(22 ) ) Since we know that cos(θ)cosec(θ) = tan(θ), on multiplication we get S = tan(20 ) tan(21 ) tan(22 )... tan 70 Lets write some more terms explicitly S = tan(20 ) tan(21 ) tan(22 )... tan(44 ) tan(45 ) tan(46 )... tan(67 ) tan(68 ) tan(69 )... cosec(70 ) Now re-write terms bef ore 45 as f ollowing S = tan(90-70 ) tan(90-69 ) tan(90-68 )... tan(90-46 ) tan(45 ) tan(46 )... tan(67 ) tan(68 ) tan(69 )... cosec(70 ) Step 5 We know that tan(90 - θ) = cot(θ) = 1/tan(θ) Now replace some of terms bef ore 45 using this equality S = [1/tan(70 )] [1/tan(69 )] [1/tan(68 )]... [1/tan(46 )] tan(45 ) tan(46 )... tan(67 ) tan(68 ) tan(69 )... cosec(70 ) Step 6 Now denominator of terms bef ore 45 will cancel with terms af ter 45. Theref ore, S = tan(45 ) S = 1 (4) 1 Expression can be rewritten as f ollowing = (cos 2 θ) 3 + (sin 2 θ) 3 + 3 cos 2 θ sin 2 θ Since x 3 + y 3 = ( x + y ) ( x 2 + y 2 - xy) = (cos 2 θ + sin 2 θ ) [ (cos 2 θ) 2 + (sin 2 θ) 2 - cos 2 θ sin 2 θ ] + 3 cos 2 θ sin 2 θ Since cos 2 θ + sin 2 θ =1 and x 2 + y 2 = (x+y) 2-2 xy) = (cos 2 θ + sin 2 θ ) [ (cos 2 θ + sin 2 θ) 2-2 cos 2 θ sin 2 θ - cos 2 θ sin 2 θ] + 3 cos 2 θ sin 2 θ = [ 1-3 cos 2 θ sin 2 θ ] + 3 cos 2 θ sin 2 θ = 1
(5) ID : in-10-trigonometry [5] BC/AB = tan 30 BC/AB = BC = AB BC = (6) 2 We need to f ind f ollowing product S = (1 + tanθ + secθ) (1 + cotθ - cosecθ) On multiplying each terms S = 1 (1 + tanθ + secθ) + cotθ (1 + tanθ + secθ) - cosecθ (1 + tanθ + secθ) S = (1 + tanθ + secθ) + (cotθ + cotθ tanθ + cotθ secθ ) - (cosecθ + cosecθ tanθ + cosecθ secθ) Using identities cotθ tanθ = 1, cotθ secθ = cosecθ and cosecθ tanθ = secθ S = (1 + tanθ + secθ) + (cotθ + 1 + cosecθ ) - (cosecθ + secθ + cosecθ secθ) Now positive cosecθ and secθ will cancel each other S = (1 + tanθ + secθ) + (cotθ + 1 + cosecθ ) - (cosecθ + secθ + cosecθ secθ) S = 2 + tanθ + cotθ - cosecθ secθ Step 5 Using identities tanθ = sinθ/cosθ, and cotθ = cosθ/sinθ sinθ S = 2 + + cosθ - cosecθ secθ cosθ sinθ S = 2 + sin2 θ + cos 2 θ sinθcosθ - cosecθ secθ S = 2 + 1 sinθcosθ - cosecθ secθ S = 2 + cosecθ secθ - cosecθ secθ S = 2
(7) b. 1 ID : in-10-trigonometry [6] S = sin 6 θ + cos 6 θ + 3sin 2 θ cos 2 θ S = (sin 2 θ) 3 + (cos 2 θ) 3 + 3sin 2 θ + 3sin 2 θ cos 2 θ S = (sin 2 θ + cos 2 θ) [(sin 2 θ) 2 - sin 2 θ cos 2 θ + (cos 2 θ) 2 ] + 3sin 2 θ cos 2 θ... Using a 3 +b 3 = (a+b)(a 2 - ab + b 2 ) S = 1[(sin 2 θ) 2 - sin 2 θ cos 2 θ + (cos 2 θ) 2 ] + 3sin 2 θ cos 2 θ... Using sin 2 θ + cos 2 θ = 1 Step 5 S = (sin 2 θ) 2 + 2sin 2 θ cos 2 θ + (cos 2 θ) 2 Step 6 S = (sin 2 θ + cos 2 θ) 2 ) 2... Using a 2 + b 2 + 2ab = (a + b) 2 Step 7 S = 1 2... Using sin 2 θ + cos 2 θ = 1 S = 1 (8) c. 26 cotθ + 5 cosec 2 θ Let, S = (cotθ + 5) (5 cotθ + 1) S = cotθ(5 cotθ + 1) + 5(5 cotθ + 1) S = 5cot 2 θ + cotθ + 5 2 cotθ + 5 S = 5(cot 2 θ + 1) + 26 cotθ Using identity cot 2 θ + 1 = cosec 2 θ S = 5(cosec 2 θ) + 26 cotθ S = 26 cotθ + 5 cosec 2 θ
ID : in-10-trigonometry [7] (9) a. 2 Lets, S = cos2 76 + cos 2 14 sin 2 76 + sin 2 14 + cos 2 14 + cos 76 sin 14 Using identities cos(θ) = sin(90 - θ) and sin(θ) = cos(90 - θ), we can re-write expression as f ollowing S = cos2 76 + sin 2 (90-14 ) sin 2 76 + sin 2 (90-14 ) + cos 2 14 + sin (90-76 ) sin 14 S = cos2 76 + sin 2 76 sin 2 76 + sin 2 76 + cos 2 14 + sin 14 sin 14 Using identity sin 2 θ + cos 2 θ = 1 S = 1/1 + cos 2 14 + sin 2 14 S = 1 + 1 S = 2 (10) c. 2sec θ S = S = S = S = S = 2sec θ
ID : in-10-trigonometry [8] (11) b. 2 Since sin(90 -θ) = cos θ and cos(90 -θ) = sin θ, expression can be rewritten as f ollowing = cos2 70 + sin 2 70 sin 2 70 + cos 2 70 + cosθ sinθ tanθ + sinθ cosθ cotθ = 1 + cos 2 θ + sin 2 θ = 1 + 1 = 2 (12) c. 14 tan θ + cot θ = 4 On squaring both sides, (tan θ + cot θ) 2 = 4 2 tan 2 θ + cot 2 θ + 2 tanθ cotθ = 16 tan 2 θ + cot 2 θ + 2 = 16 Step 5 tan 2 θ + cot 2 θ = 16-2 = 14
(13) 1 ID : in-10-trigonometry [9] It is given that 2sinθ cosθ = 2 (2sinθ cosθ) 2 = 2 2... [On squaring both sides] Now add (sinθ + 2cosθ) 2 to both sides of equation (2sinθ cosθ) 2 + (sinθ + 2cosθ) 2 = 2 2 + (sinθ + 2cosθ) 2 4sin 2 θ + cos 2 θ - 4sinθcosθ + sin 2 θ + 4cos 2 θ + 4sinθcosθ = 4 + (sinθ + 2cosθ) 2 5sin 2 θ + 5cos 2 θ + 4sinθcosθ - 4sinθcosθ = 4 + (sinθ + 2cosθ) 2 5(sin 2 θ + cos 2 θ) = 4 + (sinθ + 2cosθ) 2 5 = 4 + (sinθ + 2cosθ) 2 (sinθ + 2cosθ) 2 = 5-4 (sinθ + 2cosθ) 2 = 1 sinθ + 2cosθ = ± 1 Since 2sinθ cosθ=2, ignore the negative value. Theref ore, sinθ + 2cosθ = 1 (14) False tanθ is ratio of side opposite to angle θ to side adjacent to angle θ, in a right angle triangle. We can see that if angle θ is increased (by keeping the hypotenuse constant), side opposite to angle θ increases more rapidly than side adjacent to angle θ, theref ore value of tanθ increases. Theref ore statement is FALSE
(15) True ID : in-10-trigonometry [10] We know that tanθ is greater than cotθ, if θ > 45. Theref ore tan 87 - cot 87 will be positive, and statement "tan 87 - cot 87 > 0" is True.