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UNIT-6
TRIGONOMETRY

"The mathematician is fascinated with the marvelous beauty of the forms

he constructs, and in their beauty he finds everlasting truth."

1. If xcosθ – ysinθ = a, xsinθ + ycos θ = b, prove that x2+y2=a2+b2.

Ans: xcosθ - y sinθ = a

xsinθ + y cosθ = b
Squaring and adding
x2+y2=a2+b2.

2. Prove that sec2θ+cosec2θ can never be less than 2.

Ans: S.T Sec2θ + Cosec2θ can never be less than 2.

If possible let it be less than 2.
1 + Tan2θ + 1 + Cot2θ < 2.
 2 + Tan2θ + Cot2θ
 (Tanθ + Cotθ)2 < 2.
Which is not possible.

3. If sin = , show that 3cos-4cos3 = 0.

Ans: Sin  = ½
  = 30o
Substituting in place of  =30o. We get 0.

4. If 7sin2+3cos2 = 4, show that tan = .

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Ans: If 7 Sin2 + 3 Cos2 = 4 S.T. Tan
3
7 Sin  + 3 Cos  = 4 (Sin  + Cos )
2 2 2 2

 3 Sin2 = Cos2

Sin 2 1
 =
Cos 2 3

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 Tan2 =
3

1
Tan =
3

5. If cos+sin = cos, prove that cos - sin = sin .

Ans: Cos + Sin = 2 Cos

 ( Cos + Sin)2 = 2Cos2
 Cos2 + Sin2+2Cos Sin = 2Cos2
 Cos2 - 2Cos Sin+ Sin2 = 2Sin2 2Sin2 = 2 - 2Cos2
 (Cos - Sin)2 = 2Sin2 1- Cos2 = Sin2 & 1 - Sin2 = Cos2
or Cos - Sin = 2 Sin.

6. If tanA+sinA=m and tanA-sinA=n, show that m2-n2 = 4

Ans: TanA + SinA = m TanA – SinA = n.

2 2
m -n =4 mn .
m2-n2= (TanA + SinA)2-(TanA - SinA)2
= 4 TanA SinA
RHS 4 mn = 4  TanA  SinA (TanA  SinA)
= 4 Tan 2 A  Sin 2 A
Sin 2 A  Sin 2 ACos 2 A
=4
Cos 2 A
Sin 4 A
=4
Cos 2 A
Sin 2 A
=4 = 4 TanA SinA
Cos 2 A
m2 – n2 = 4 mn

7. If secA= , prove that secA+tanA=2x or .

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Ans: Sec = x +
4x
1 2
 Sec2 =( x + ) (Sec2= 1 + Tan2)
4x
1 2
Tan2 = ( x + ) -1
4x
1 2
Tan2 = ( x - )
4x

1
Tan = + x -
4x

1 1
Sec + Tan = x + + x-
4x 4x
1
= 2x or
2x

8. If A, B are acute angles and sinA= cosB, then find the value of A+B.

Ans: A + B = 90o

9. a)Solve for , if tan5 = 1.

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Ans: Tan 5 = 1 =  =9o.
5

Sin 1  Cos
b)Solve for  if 1  Cos  Sin
 4.

Sin 1  Cos
Ans:  4
1  Cos Sin

Sin 2   1(Cos) 2
4
Sin(1  Cos)

Sin 2   1  Cos 2   2Cos

4
Sin  SinCos

2  2Cos
4
Sin (1  Cos )

2  (1  Cos )
 4
Sin (1  Cos )

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2
 4
Sin

1
 Sin =
2
 Sin = Sin30
 = 30o

10. If

Cos Cos
Ans: m n
Cos Sin

Cos 2 Cos 2
m 2=
n 2=
Cos 2  Sin 2 

LHS = (m2+n2) Cos2 

 Cos 2 Cos 2 
 Cos 
2

 Cos  Sin  
2 2

 1 
Cos 2  Cos 2 
 Cos Sin 
2 2
= 

Cos 2
= =n2
Sin 2 
(m2+n2) Cos 2  =n2

11. If 7 cosec-3cot = 7, prove that 7cot - 3cosec = 3.

Ans: 7 Cosec-2Cot=7
P.T 7Cot - 3 Cosec=3
7 Cosec-3Cot=7
7Cosec-7=3Cot
7(Cosec-1)=3Cot
7(Cosec-1) (Cosec+1)=3Cot(Cosec+1)
7(Cosec2-1)=3Cot(Cosec+1)
7Cot2=3 Cot (Cosec+1)
7Cot= 3(Cosec+1)

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7Cot-3 Cosec=3

12. 2(sin 6+cos6) – 3(sin4+cos4)+1 = 0

Ans: (Sin2)3 + (Cos2)3-3 (Sin4+(Cos4)+1=0

Consider (Sin2)3 +(Cos2)3
(Sin2+Cos2)3-3 Sin2Cos2( Sin2+Cos2)
= 1- 3Sin2 Cos2
Sin4+Cos4(Sin2)2+(Cos2)2
= (Sin2+Cos2)2-2 Sin2 Cos2
= 1- 2 Sin2 Cos2
= 2(Sin6+Cos6)-3(Sin4+Cos4) +1
= 2 (1-3 Sin2 Cos2)-3 (1-2 Sin2+Cos2)+1

13. 5(sin8A- cos8A) = (2sin2A – 1) (1- 2sin2A cos2 A)

Ans: Proceed as in Question No.12

5
14. If tan = &  + =90o what is the value of cot.
6
Ans: Tan = 5 6 i.e. Cot = 5 6 Since  +  = 90o.

15. What is the value of tan in terms of sin.

Sin
Ans: Tan  = Cos
Sin
Tan  =
1  Sin 2

16. If Sec+Tan=4 find sin , cos

Ans: Sec  + Tan  = 4

1 Sin =4

Cos Cos

1  Sin
4
Cos
(1  Sin ) 2
  16
Cos 2

 apply (C & D)

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(1  Sin ) 2  Cos 2 16  1
= =
(1  Sin ) 2  Cos 2 16  1

2(1  Sin ) 17
 2 Sin (1  Sin ) =
15
1 17
 Sin =
15
15
Sin=
17
Cos = 1  Sin 2
2
 15  8
1   =
 17  17

p2 1
17. Sec+Tan=p, prove that sin =
p2 1

P2 1
Ans: Sec + Tan= P. P.T Sin=
P2 1
Proceed as in Question No.15

18. Prove geometrically the value of Sin 60o

Ans: Exercise for practice.

1  tan  3 1 sin 
19. If  ,show that =1
1  tan  3 1 cos 2
Ans: Exercise for practice.

2  2 1 
20. If 2x=sec and = tan ,then find the value of 2  x  2  . (Ans:1)
x  x 
Ans: Exercise for practice.

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