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# SSC CGL Mains Trigonometry Questions : 29th July

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As you all know, the upcoming months are lined up with various important exams like SSC CGL Mains 2018, so we are here to help you with the subject that is common to all of the given exams. We are providing daily quantitative aptitude quizzes, practice which will help you to score good marks in this section. We aim to provide the best study material to our readers with exam level questions to help them get used to the recent pattern. Attempt this quiz and check your preparation.

S1. Ans. (b)
Sol.
sin⁡x/sin⁡〖y 〗 =p & cos⁡x/cos⁡y =q
sin⁡x/p=sin⁡y
⇒sin^2⁡x/p^2 =sin^2⁡y
cos⁡x/q=cos⁡〖y⇒cos^2⁡x/q^2 =cos^2⁡y 〗 , Adding both equation
sin^2⁡x/p^2 +cos^2⁡x/q^2 =sin^2⁡〖y+cos^2⁡〖y=1〗 〗 divide by cos^2⁡x
tan^2⁡x/p^2 +1/q^2 =1/cos^2⁡x
tan^2⁡〖x(1/p^2 -1)=1-1/q^2 〗 ,tan^2⁡〖x=((q^2-1)/q^2 )/((1-p^2)/p^2 )〗 tan⁡〖x=p/q √((q^2-1)/(1-p^2 ))〗

Q2. If 2 sin⁡〖θ+cos⁡〖θ=7/3〗,〗 then the value of (tan^2⁡〖θ-sec^2⁡θ 〗 ) ?
यदि 2 sin⁡〖θ+cos⁡〖θ=7/3〗,〗 तो (tan^2⁡〖θ-sec^2⁡θ 〗 ) का मान ज्ञात करें
(a) 7/3
(b) 0
(c) -1
(d) 3/7

S2. Ans.(c)
Sol.
tan^2⁡〖θ-sec^2⁡θ 〗
=-(sec^2⁡〖θ-tan^2⁡θ 〗 )=-1

Q3. What is the value of [(sec 2θ+1) √(sec^2 θ–1)]×1/2 (cot θ–tan θ)?
[(sec 2θ+1) √(sec^2 θ–1)]×1/2 (cot θ–tan θ) का मान क्या है?
(a) 0
(b) 1
(c) cosec θ
(d) sec θ

S3. Ans.(B)
Sol. GIVEN
((sec⁡2θ+1) √(sec^2⁡θ–1))×1/2 (cot⁡θ–tan⁡θ )
[(sec⁡2θ+1) tan⁡θ ]×1/2 (cot⁡θ–tan⁡θ )
[(1/cos⁡2θ +1) tan⁡θ ]×1/2 (cot⁡θ–tan⁡θ )
[((1+2 cos^2⁡θ–1) tan⁡θ)/(2 cos^2⁡θ–1)]×1/2 (cot⁡θ–tan⁡θ )
[(2 cos⁡θ sin⁡θ)/(2 cos^2⁡θ–1)]×1/2 [cos⁡θ/sin⁡θ –sin⁡θ/cos⁡θ ]
= [sin⁡2θ/cos⁡2θ ×1/2 cos⁡2θ/(sin⁡θ cos⁡θ )]=1

Q4. What is the value of sin (630° + A) + cos A?
sin (630° + A) + cos A का मान क्या है?
(a) √3/2
(b) 1/2
(c) 0
(d) 2/√3

S4. Ans.(C)
Sol.
〖 sin〗⁡(630+A)+cos⁡A
⇒ sin⁡[2×360 –(90 –A)]+cos⁡A
⇒ –sin⁡(90–A)+cos⁡A
⇒ –cos⁡A+cos⁡A⇒0

Q5. If tan α = n tan β and sin α = m sin β, then (m^2-1)/(n^2-1)= ?
यदि tan α = n tan β और sin α = m sin β, तो (m^2-1)/(n^2-1)= ?
(a) cos³ α
(b) sin³ α
(c) sin² α
(d) cos² α

.

S5. Ans.(d)
Sol.
m^2–1=(sin^2⁡α–sin^2⁡β)/sin^2⁡β
n^2–1=(tan^2⁡α–tan^2⁡β)/tan^2⁡β
=(sin^2⁡α cos^2⁡β–sin^2⁡β.cos^2⁡α)/(cos^2⁡α cos^2⁡β )×cos^2⁡β/sin^2⁡β
=(sin^2⁡α (1–sin^2⁡β )–sin^2⁡β (1–sin^2⁡α ))/(sin^2⁡β.cos^2⁡α )
=(sin^2⁡α–sin^2⁡β)/(sin^2⁡β.cos^2⁡α )
∴(m^2–1)/(n^2–1)=(sin^2⁡α–sin^2⁡β)/sin^2⁡β ×(sin^2⁡β cos^2⁡α)/(sin^2⁡α–sin^2⁡β )
=cos^2⁡α

Q6. What is the value of [(sin 59° cos 31° + cos 59° sin 31°)÷(cos 20° cos 25° – sin 20° sin 25°)]?
[(sin 59° cos 31° + cos 59° sin 31°)÷(cos 20° cos 25° – sin 20° sin 25°)] का मान क्या है?
(a) 1/√2
(b) 2√2
(c) √3
(d) √2

S6. Ans.(D)
Sol.
(sin⁡59 cos⁡31+cos⁡59 sin⁡31)/(cos⁡20 cos⁡25–sin⁡20 sin⁡25 )
sin⁡(A+B)=sin⁡A cos⁡B+cos⁡A sin⁡B ——FORMULA
cos⁡(A+B)=cos⁡A cos⁡B – sin⁡A sin⁡B —– FORMULA
⇒ sin⁡(59+31)/cos⁡(20+25) =sin⁡90/cos⁡45 =√2

Q7. The eliminate of θ from x cos⁡θ-y sin⁡θ=2 and x sin⁡θ+y cos⁡θ=4 will give-
x cos⁡θ-y sin⁡θ=2 और x sin⁡θ+y cos⁡θ=4 में θ को हटाने के बाद मान ज्ञात करें
(a) x^2+y^2=20
(b) 3x^2+y^2=20
(c) x^2-y^2=20
(d) 3x^2-y^2=10

S7. Ans.(a)
Sol. x cos⁡θ-y sin⁡θ=2
x sin⁡θ+y cos⁡θ=4
x^2 〖cos〗^2⁡θ+y^2 〖sin〗^2⁡θ-2xy sin⁡θ.cos⁡θ+x^2 〖sin〗^2⁡θ+y^2 〖cos〗^2⁡θ+2xy sin⁡θ.cos⁡θ =4+16 =20
x^2+y^2=20

Q8. ( sec^2⁡θ-cot^2⁡(90-θ))/(cosec^2⁡67-tan^2⁡23 )+sin^2⁡40+sin^2⁡50 is equal to
(sec^2⁡θ-cot^2⁡(90-θ))/(cosec^2⁡67-tan^2⁡23 )+sin^2⁡40+sin^2⁡50 किस के बराबर है?
(a) 0
(b) 4
(c) 2
(d) 1

S8. Ans.(c)
Sol. (〖sec〗^2⁡θ-〖cot〗^2⁡(90-θ))/(〖cosec〗^2⁡67-〖tan〗^2⁡23 )+〖sin〗^2⁡40+〖sin〗^2⁡50

=(〖sec〗^2⁡θ-〖tan〗^2⁡θ)/(〖cosec〗^2⁡67-〖cot〗^2⁡67 )+〖sin〗^2⁡40+〖cos〗^2⁡〖40 〗
=1+1
=2

Q9. What is the value of [sin (90° – 10θ) – cos (π – 6θ)]/ [cos (π /2 – 10θ) – sin (π – 6θ)]?
[sin (90° – 10θ) – cos (π – 6θ)]/ [cos (π /2 – 10θ) – sin (π – 6θ)] का मान क्या है?
(a) tan 2θ
(b) cot 2θ
(c) cot θ
(d) cot 3θ

S9. Ans.(B)
Sol. (sin⁡(90–10θ)–cos⁡(π–6θ))/(cos⁡(π/2–10θ)–sin⁡(π–6θ))
⇒ (cos10θ+cos6θ)/(sin10θ–sin6θ) (AS sin⁡(90–10θ)=cos⁡〖10θ &cos⁡(π–6θ)=-cos⁡6θ 〗)(apply{ cos a+cos b}&{sin a –sin b} formula )
⇒(2cos8θ×cos2θ)/(2cos8θ×sin2θ)⇒▭cot2θ

Q10. If sec θ (cos θ + sin θ) = √2, then what is the value of (2 sin θ)/(cos θ – sin θ)?
यदि sec θ (cos θ + sin θ) = √2, तो (2 sin θ)/(cos θ – sin θ) का मान क्या है?
(a) 3√2
(b) 3/√2
(c) 1/√2
(d) √2

S10.Ans(D)
Sol. secθ (cos⁡〖θ+sinθ)〗=√2To find =(2 sinθ)/(cosθ-sinθ)

cosθ + sinθ = √2 cosθ…………(1)
squaring both sideseq 1
cos2θ + sin2θ + 2cosθ sinθ = 2cos2θ
2 sinθ cosθ = cos2θ–sin2θ
(2 sinθ)/(cosθ–sinθ) = ((cosθ + sinθ))/cosθ (from equation 1….((cosθ + sinθ))/cosθ= √2 )
= √2

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