Tuesday 27 September 2011

CBSE Maths X Ch-6 : Trigonometry Identities

1 mark questions
Q. 1 Write the value of sin 620 sin 280 – cos620 cos 280
Q. 2 Write  cot in terms of sin A.
Q. 3 Express sec790 + cot 610 in terms of trigonometrical ratios of angles between 00and 450 .
Q. 4 If 3tanθ = 4 , then write the value of tan θ + cot θ .
Q. 5 If sinq – cos θ = 0 , 00 <θ < 900 , then write the value of 'θ ' .
Q. 6 If 'q ' , then write the value of sin θ + cos 2θ .
Q. 7 Write the value of sin2 740 + sin2 160 .
Q. 8 In ΔABC, = 900 and sin C = 4/5 , what is the value of cos A?
Q. 9 If A and B are acute angles and sin = cos   , than write the value of A+B.
Q. 10 Write the value of tan2 300 + sec2 450 .
Q. 11 Write the value of 9 cosec2620 – 9 tan2 280 .
Q. 12 If sin q = 1/2, write the value of sin q – cosec θ .
Q. 13 What is the value of cos2490 – sin2 410 ?
Q. 14 If q = 450  , then what is the value of 2cos ec2θ + 3sec2q ?
Q. 15 Write the value of sin ( 900 –q ) cos q + cos(900 – q) Sinq
Q. 16 If tan (3–150 ) =1, than write the value of 'x'.
Q. 17 In ΔABC, write tan (AB)/2  in terms of angle 'C'.
Q. 18 If q = 300 , then write the value of 1 – tan2 2q .
Q. 19 If tanq + cotq = 3, then what is the value of tan2θ + cot2θ ?
Q 20 Write the value of cot (35+q ) – tan (550 – q )
2/3 marks questions
Q. 21 If sin 2q = cos (q – 36)0, 2θ and (q – 360) are acute angles. Find the value of 'q ' .
Q. 22 If tan (320 +q ) = cotq , θ and (320 +q ) are acute angles, find the value of 'q '.
Q. 23 If sin ( AB) =1 and cos(A – B) = √3 /2 ,  00 ≤ ( AB) ≤ 900 , A > B, then find the values of A and B.
Q. 24 If q = 300 , then find the value of (1– tan2q)/(1+ tan2q)
Q. 25 If tan q = √2 –1, then find the value of (2 tanq) /(1+ tan2q)
Q. 26 If q = 300 , then verify : cos 3q = 4cos3q – cosq .
Q. 27 Simplify: tan 2 600 + 4cos2 450 + 3sec2 300 + 5cos2 900
Q. 28 Find the value of :-
(sin 620)/ cos 28  +  2 (tan 730)/ cot17–  (2sin 28 .sec62)/(cot170 7sec 320 – 7cot 580)
Q. 29 find the value of   
(11 sin 700 ) / ( 7 cos200 ) –  (4/7) [(cos530 .cos 370) / (tan150 .tan350 tan550 tan750)
Q. 30 Find the value of :-
3(sin2 740 sin216 )/( 4sin 620 .sec280) + 3(tan2 280 – cosec 2620 )/ tan 250 .tan 350tan 550 tan 650)

1 comment:

  1. Nice idea!! Thank you so much for such information. I’ve been waiting patiently for your next blog entry!
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