Q1.If sinθ + cosecθ = 3,then find the value of sin4 θ +1 sin2 θ [Ans:- 7] Q2 .If tanθ = pq ,show that p sinθ – q cosθ p2- q2 p sinθ + q cosθ p2+ q2 Q3 If a cos θ – b sin θ=x and a sin θ+b cos θ = y prove that a2+b2 = x2+y2 Q4.If sin θ and sec θ, ( 00 < θ < 900) are the roots of the equation √3x2+kx+3=0,then find the value of k [Ans:- k=- 3+4√3 2 ] Q5. Show that geometrically that sin 600 =√32 Q6. A rhombus of side 20cm has two angles of 600 each.Find the length of the diagonals. Q7.Prove that sec6θ =tan6θ +3tan2θsec2θ+1 Q8. If xa cos θ+ ybsinθ=1 and xa sin θ- ybcosθ=1 prove that x2 y2 2 a2 b2
Answers
ANSWER
8.x2+αy2+2βy=a2 represent pair of perpendicular straight lines
Coeff. of x2+Coeff. of y2=0
1+α=0
α=−1
and Discriminant=D=abc+2fgh−af2−bg2−ch2=0
=−αa2−β2=0
αa2+β2=0
and α=−1,β2=a2
α−1 and β=a or β=
7.According to problem,
sec6θ=(sec2θ)3=(1+tan2θ)3=(1)3+(tan2θ)3+3×1×tan2θ(1+tan2θ)=1+tan6θ+3tan2θsec2θ
Hence proved
6.Let ABCD be a rhombus of side 10cm and ∠BAD=∠BCD=60o. Diagonals of parallelogram bisect each other.
So, AO=OC and BO=OD
In right triangle AOB
sin30o=ABOB
⇒ 21=10OB
⇒ OB=5cm
∴ BD=2(OB)
⇒ BD=2(5)
⇒ BD=10cm
cos30o=ABOA
⇒ 23=10OA
⇒ OA=53
∴ AC=2(OA)
⇒ AC=2(53)
⇒ AC=103cm
So, the length of diagonals AC=103cmand BD=
5.∫1+sin2x1−sin2xdx
=∫sin2x+cos2x+2sinxcosxsin2x+cos2x−2sinxcosxdx
=∫(sinx+cosx)2(sinx−cosx)2dx
=∫sinx+cosxsinx−cosxdx
Let t=sinx+cosx⇒dt=(cosx−sinx)dx=−(sinx−cosx)dx
=−∫tdt
=−log∣t∣+c
=−log∣sinx+cosx∣+c where t=sinx+cosx
4.⇒ The given quadratic equation is 2x2+kx+4=0, comparing it with ax2+bx+c=0
⇒ We get, a=2,b=k,c=−4
⇒ It is given that roots are real.
∴ b2−4ac≥0
⇒ (k)2−4(2)(4)≥0
⇒ k2−32≥0
⇒ (k+42)(k−42)≥0
⇒ k≤−42 or k≥42
3.
We have: cos2θaxsinθ−sin2θbycosθ=0
⇒axsin3θ−bycos3θ=0
⇒bysin3θ=axcos3θ
⇒(bysin3θ)2/3=(axcos3θ)2/3
⇒(by)2/3sin2θ=(ax)2/3cos2θ
⇒(by)2/3sin2θ−(ax)2/3cos2θ−(by)2/3
2.answr
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MATHS
Asked on December 20, 2019 byAnshuli Bhadwa
If qcosΘ=q2−p2, prove that q sin Θ=p.
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ANSWER
Given:- qcosθ=q2−p2
To prove:- qsinθ=p
Proof:-
qcosθ=q2−p2(Given)
⇒cosθ=qq2−p2.....(1)
As we know that,
cosθ=HypotenuseBase.....(2)
On comparing eqn(1)&(2), we have
Base=q2−p2
Hypotenuse=q
Now applying pythagoras theorem,
Hypotenuse2=Base2
1.cscθ−sinθ=a3
sinθ1−sinθ=a3
sinθ1−sin2θ=a3
a3=sinθcos2θ=cotθcosθ ....... (i)
Also, secθ−cosθ=b3
cosθ1−cosθ=b3
b3=cosθ1−cos2θ=cosθsin2θ
b3=tanθsinθ ....... (ii)
Consider,
a2b2(a2+b2)=[cotθcosθtanθsinθ]2/3[a2+b2]
=[sin
Explanation:
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