Math Chapter Wise

EnglishNVS PGT 10-06-2019

Q81.sin7x 6sin5x 17sin3x 12sinx sin6x 5sin4x 12sin2x + + + + + + + + + + + + + + equals:

Acos x B2 cos x Csin x D2 sin x

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EnglishKVS PGT 23-12-2018 SECTION B PGT TIER-I 31.11.2014

Q82.If sinθ + cosθ = 1, then the value of sin2θ is:

A1 B1/2 C0 D–1

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EnglishKVS PGT 23-12-2018

Q83.The general solution of the equation 2 sin sec 3 tan 0 θ + θ = θ + θ = θ + θ = θ + θ = is:

A()n 1 n 1. 3 + π θ = π + − Bn θ = π C()n 1 n 1. 6 + π θ = π + − Dn 2 π θ = Where n∈Z

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EnglishUP PGT-2016 (02-02-2019)

Q84.The equation 4 2 2 sin 2cos a 0 θ − θ + = θ − θ + = θ − θ + = θ − θ + = will have atleast one solution, if

A1 a 1 2 ≤ Ba 2/≤ Ca 2/≤ Da 1/≥

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EnglishDSSSB 23 SEP 2018 (9 AM-11 AM)

Q85.x for general value for which given as 2 a sinx cosx min 1,a 4a 7 →π + = − + + = − + + = − + + = − + went:

A() { } n n 1 1 4 π + − − B() n 2n 1 2 π + − C2n 4 π + D() n n 1 2 π + −

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EnglishDSSSB 23 SEP 2018 (9 AM-11 AM)

Q86.Solution which of the given relation is true?

Acos1o = cos1 Bcos1o/> cos1 Co cos1 sin1 180 π = Dcos1o/< cos1

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EnglishDSSSB 23 SEP 2018 (9 AM-11 AM)

Q87.The distance of the highest point on the graph of the function y 3cosx sin x = + from the x- axis is.

A2 B1/2 C3/2 D2/3

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EnglishDSSSB TGT (SECTION-B) 28 DEC 2014

Q88.3 tan 4 and θ lies in the second quadrant, then sin θ is equal to–

A2/5 B2/3 C3/5 D4/5

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EnglishDSSSB TGT (SECTION-B) 28 DEC 2014

Q89.The value of 13 sin 4 is same as the value of–

A3 sin 4 B5 sin 4 Csin/4 D9 sin 4

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EnglishDSSSB TGT (SECTION-B) 28 DEC 2014

Q90.If sin (A + B) = cos (A – B) 3 2 acute angles of A and B are–

A450/and 150 B300/and 600 C150/and 600 DNone of these

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EnglishJDD-75-PGT TIER II-X-15 28.06.2015

Q91.If 15 sin 17 θ = θ = then the value of cosθ and tanθ are-

A8 15, 15 8 − B8 15, 17 8 C8 15, 17 8 − D15 15, 17 8

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EnglishJDD-75-PGT TIER II-X-15 28.06.2015

Q92.If 2 2 2 cos A cos C sin B + = + = then triangle ABC is–

Aright angle Bequilateral Cisosceles Dnone of these

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EnglishJDD-75-PGT TIER II-X-15 28.06.2015

Q93.ln a triangle ABC, a =13, b=14, c =15. Then sin A 2 is equal to–

A1/5 B2/5 C3/5 D4/5

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EnglishJDD-75-PGT TIER II-X-15 28.06.2015

Q94.lf cos A cosB cosC a b c =, then triangle ABC is–

Aisosceles Bright angled Cequilateral Dno conclusion

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EnglishJDD-75-PGT TIER II-X-15 28.06.2015

Q95.The sides of a triangle are in the ratio 1: 3: 2, then the angles of the triangle are in the ratio–

A1:3: 5 B2: 3: 4 C3: 2:1 D1: 2:3

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EnglishTGT 2003

Q96.If x=cos10° ° ° cos20° ° ° cos40° ° °, then the value of x is:

A1 8 tan 10° B1 8 cosec10° C1 8 cot 10° D1 8 sec10°

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EnglishTGT 2001

Q97.It the area of a triangle is 81 cm: and its semi perimeter is 27 cm, then the radius of the in circle will be:

A6 cm. B3 cm. C1.5 cm. Dnone of these

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EnglishTGT 2001

Q98.The value of sin cot–1 x is:

A2 1 x + Bx C(1+x) –3/2 D(1+x2/) –1/2

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EnglishTGT 2001

Q99.The general solution of the equation 2 tan 2 3 tan 1 α + α = α + α = α + α = α + α = is:

A2 π θ = B1 n 2 + π C() 6n 1 12 π + Dn 12 π

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EnglishTGT 2001

Q100.The value of 0 0 cot40 1 cos35 tan50 2 sin55 − is:

A1 B0 C1/2 D1/2

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