SSC CGL Mains Trigonometry Questions : 29th July_00.1
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SSC CGL Mains Trigonometry Questions : 29th July

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SSC CGL Mains Trigonometry Questions : 29th July_50.1

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
On squaring and adding
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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