Grade 10 Trigonometry

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1 ID : pk-0-trigonometry [] Grade 0 Trigonometry For more such worksheets visit Answer t he quest ions () Simplif y - 2 sin 3 θ - 2 cos 3 θ (2) If secθ tan θ y, simplif y in terms of θ. (3) [sec(7 ) sec(8 ) sec(9 )... sec(83 ) ] [sin(7 ) sin(8 ) sin(9 ) ]? (4) Simplif y ( cotθ - cosecθ) ( tanθ secθ) (5) Simplif y cos 6 θ sin 6 θ 3 cos 2 θ sin 2 θ (6) Simplif y cot 2 β cosecβ. (7) Simplif y (8) Simplif y 3(sin 4 θ cos 4 θ) - 2(sin 6 θ cos 6 θ) Choose correct answer(s) f rom given choice (9) (cosecθ ) 2 (secθ ) 2? a. ( sin θ cos θ) 2 b. ( sec θ cosec θ) 2 c. ( - sin θ cos θ) 2 d. (sec θ cosec θ) 2 (0)? a. 2-2 sin 2 θ b. 2 2 sin 2 θ - c. 2 - cos 2 θ d. 2 2 sin 2 θ () Simplif y a. sin 2 θ - cos 2 θ b. c. d. 0

2 ID : pk-0-trigonometry [2] (2) If - 0, f ind value of tan 4 θ - sin 4 θ. a. 0.5 b c d. (3) Simplif y. a. tanθ cotθ b. sin 2 θ - cos 2 θ c. - d. tanθ cotθ (4) Simplif y a. cos 2 θ - sin 2 θ b. c. - d. tanθ - cotθ Check True/False (5) The angle of elevation of the top of a pole is 30. If the height of the pole is tripled, then the angle of elevation of its top will also be tripled. True False 206 Edugain ( All Rights Reserved Many more such worksheets can be generated at

3 Answers ID : pk-0-trigonometry [3] () -tan θ Let, S - 2 sin 3 θ - 2 cos 3 θ On taking and common, S S ( - 2 sin 2 θ) ( - 2 cos 2 θ) [( - sin 2 θ) - sin 2 θ] [( - cos 2 θ) - cos 2 θ] Using identity sin 2 θ cos 2 θ [cos 2 θ - sin 2 θ] S [sin 2 θ - cos 2 θ] S -tan θ (2)

4 (3) ID : pk-0-trigonometry [4] We need to f ind some of f ollowing series S (sec(7 ) sec(8 ) sec(9 )... sec(83 ) )(sin(7 ) sin(8 ) sin(9 ) ) Since we know that sec(θ)sin(θ) tan(θ), on multiplication we get S tan(7 ) tan(8 ) tan(9 )... tan 83 Lets write some more terms explicitly S tan(7 ) tan(8 ) tan(9 )... tan(44 ) tan(45 ) tan(46 )... tan(80 ) tan(8 ) tan(82 )... sin(83 ) Now re-write terms bef ore 45 as f ollowing S tan(90-83 ) tan(90-82 ) tan(90-8 )... tan(90-46 ) tan(45 ) tan(46 )... tan(80 ) tan(8 ) tan(82 )... sin(83 ) Step 5 We know that tan(90 - θ) cot(θ) /tan(θ) Now replace some of terms bef ore 45 using this equality S [/tan(83 )] [/tan(82 )] [/tan(8 )]... [/tan(46 )] tan(45 ) tan(46 )... tan(80 ) tan(8 ) tan(82 )... sin(83 ) Step 6 Now denominator of terms bef ore 45 will cancel with terms af ter 45. Theref ore, S tan(45 ) S

5 (4) 2 ID : pk-0-trigonometry [5] We need to f ind f ollowing product S ( cotθ - cosecθ) ( tanθ secθ) On multiplying each terms S ( tanθ secθ) cotθ ( tanθ secθ) - cosecθ ( tanθ secθ) S ( tanθ secθ) (cotθ cotθ tanθ cotθ secθ ) - (cosecθ cosecθ tanθ cosecθ secθ) Using identities cotθ tanθ, cotθ secθ cosecθ and cosecθ tanθ secθ S ( tanθ secθ) (cotθ cosecθ ) - (cosecθ secθ cosecθ secθ) Now positive cosecθ and secθ will cancel each other S ( tanθ secθ) (cotθ cosecθ ) - (cosecθ secθ cosecθ secθ) S 2 tanθ cotθ - cosecθ secθ Step 5 Using identities tanθ /, and cotθ / S 2 - cosecθ secθ S 2 sin2 θ cos 2 θ - cosecθ secθ S 2 - cosecθ secθ S 2 cosecθ secθ - cosecθ secθ S 2

6 (5) ID : pk-0-trigonometry [6] 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 and x 2 y 2 (xy) 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 [ - 3 cos 2 θ sin 2 θ ] 3 cos 2 θ sin 2 θ (6) cosec β Let, S cot 2 β cosecβ Using identity cot 2 θ cosec 2 θ -, S cosec2 β - cosecβ Using a 2 - b 2 (ab)(a-b), (cosecβ )(cosecβ - ) S cosecβ S (cosecβ - ) S cosecβ

7 ID : pk-0-trigonometry [7] (7) sin 2 θ cos 2 θ We have been asked to simplif y the ( cotθ tanθ)( - ) sec 3 θ - cosec 3 θ. ( cotθ tanθ)( - ) sec 3 θ - cosec 3 θ ( )( - ) (secθ - cosecθ)(sec 2 θ secθ cosecθ cosec 2 θ) ) [Since, a 3 - b 3 (a - b)(a 2 ab b 2 ] ( ( cos2 θ sin 2 θ )( - ) - )( cos 2 θ cos 2 θ ) [Since, secθ and cosecθ ] ( )( - ) ( - )(sin 2 θ cos 2 θ) ( )(sin 2 θ cos 2 θ) ( )( - )(sin3 θ cos 3 θ) ( )( - )( ) sin 2 θ cos 2 θ Theref ore, ( cotθ tanθ)( - ) sec 3 θ - cosec 3 θ is equal to the sin 2 θ cos 2 θ.

8 (8) ID : pk-0-trigonometry [8] 3 (sin 4 θ cos 4 θ) - 2 [ (sin 2 θ) 3 (cos 2 θ) 3 } ] 3 (sin 4 θ cos 4 θ) - 2 [ (sin 2 θ cos 2 θ) {(sin 2 θ) 2 (cos 2 θ) 2 - sin 2 θ cos 2 θ } ] Since sin 2 θ cos 2 θ 3 (sin 4 θ cos 4 θ) - 2 {sin 4 θ cos 4 θ - sin 2 θ cos 2 θ } sin 4 θ cos 4 θ 2 sin 2 θ cos 2 θ Step 5 (sin 2 θ cos 2 θ) 2 2 (9) b. ( sec θ cosec θ) 2 (0) a. 2-2 sin 2 θ On adding two f ractions, S Let, S S Using identity sin 2 θ cos 2 θ, S S S S

9 () d. 0 ID : pk-0-trigonometry [9] On adding two f ractions (2) c It is given that - 0, which means that From above relation we can derive that tanθ We also know that, sin 2 θ /() sin 2 θ /2 Theref ore, tan 4 θ - sin 4 θ tan 4 θ - sin 4 θ 0.75

10 (3) a. tanθ cotθ ID : pk-0-trigonometry [0] We know that, tanθ, cotθ Now, cotθ - tanθ tanθ - cotθ can be simplif ied as: cotθ - tanθ tanθ - cotθ cos 2 θ ( - ) sin 2 θ ( - ) cos 3 θ - sin 3 θ.( - ) ( - )(cos2 θ. sin 2 θ).( - ) cos2 θ. sin 2 θ. cos 2 θ... sin 2 θ. cotθ tanθ tanθ cotθ

11 (4) b. ID : pk-0-trigonometry [] Multiply numerator and denominator of f irst term by Since tanθ / (5) False