“Arise! Awake! Stop not till the Goal is reached” 1 /6
PROBLEMS on TRIGONOMETRY
1.
(With Hints/Solutions)
Prove that,
(i) tan 70 = tan 20 + 2 tan50. (ii) tan 50 = tan 40 + 2.tan 10 .
(iii) tan 70 = 2tan40 + tan20 + 4.tan10.
[HINTS: (i) Since 70 + 20 = 90
Now, tan(70  20 ) 
tan 70  tan 20
1  tan 70.tan 20
 tan 70  tan 20  2.tan(70  20 )
 tan 70  tan 20  2.tan 50
,
[Note that,
If A + B = 90,
then tan(A  B)  tan A  tan B
 tan A  tan B  2.tan(A  B) this is an identity
1  tan A.tan B
Use this identity, to prove such type of problems. Choose values of A & B such that A + B = 90. In
such type of problems value of (A – B) (here 50) will also be present. ]
[HINTS: (ii) Let A = 50° and B = 40°, A + B = 90°
Then, tan A  tan B  2.tan(A  B)  tan 50  tan 40  2.tan10  tan 50  tan 40  2.tan10 ]
[HINTS: (iii) If A + B = 90°, then
tan A  tan B
tan(A  B) 
 tan A  tan B  2.tan(A  B)  tan A  tan B  2.tan(A  B) (i)
1  tan A.tan B
In (i), putting, A=70°, B=20°,
tan 70  tan 20  2.tan 50 (ii)
tan 50  tan 40  2.tan10  (iii)
Again In (i), putting, A=50°, B=40°,
By (ii) & (iii), tan 70  tan 20  2.(tan 40  2 tan10 )  tan 70  2 tan 40  tan 20  4 tan10 (proved) ]
2.
Prove that, (i) cot 70 + 4 cos70 = 3
(iii) 4.sin 50 – 3.tan 50 = 1.
[HINTS: (i) LHS  cot 70  4cos 70 
cos 70  cos 50  cos 50

sin 70

sin 70
cos 70  2.2cos 70.sin 70
sin 70

cos 70  2sin140
sin 70
[ sin140  sin(90  50 )  cos 50 ]

2cos 60.cos10  cos 50
(ii) tan 70 + tan 10 – tan50 = 3.
(iv) tan 20 + tan 70 = 2.cosec 40.

cos10  cos 50
sin 70

2 cos 30. cos 20
cos 20
2
3
 3  RHS ]
2
[HINTS: (ii) LHS  tan 70  tan10  tan 50  tan(60  10 )  tan(60  10 )  tan10

tan 60  tan10
tan 60  tan10
3  tan10
3  tan10



tan10


 tan10






1  tan 60 . tan10 1  tan 60 . tan10
1  3 tan10 1  3 tan10




 
3  tan10 1  3 tan10 
2

3  tan10 1  3 tan10

1  3 tan 10
3.2 3 tan10  2 tan10  tan10  3 tan 3 10
 (0342) 262 4499
1  3tan 2 10
 3.
  tan10
3 tan10  tan3 10
1  3 tan 2 10
 3.tan 30  3  RHS ]
CONFI-HS-PR&SOL-TRIG-02-161304
A Mathematics Center for Excellence
“Arise! Awake! Stop not till the Goal is reached” 2 /6
[HINTS: (iii) LHS  4sin 50  3 tan 50 
2.2 sin 50.cos 50  3 sin 50
3
sin 50
2
cos 50
2 sin100  2

cos 50
2 cos10  2 sin 60 sin 50 2 cos10  cos10  cos110 cos10  cos110



cos 50
cos 50
cos 50
2 cos 60 cos 50

 1  RHS ]
cos 50
1
2
[HINTS:(iv) LHS  tan 20  tan 70  tan 20  cot 20 

 2 cos ec40




sin 20 cos 20
3.
2 sin 20 cos 20
2
If tan(A  B)  sin C  1 , prove that, tanA.tanB = tan2C.
tan A
sin 2 A
2
2
[HINTS: tan(A  B)  sin C  1  sin C  sin A cos(A  B)  sin(A  B) cos A
2
2
tan A
sin A
sin( A  A  B)
 sin 2 C  sin A sin 2 A
sin A cos(A  B)
sin 2 C


2
1  sin C
4.
sin Asin B
cos(A  B)  sin A sin B
2

sin C sin A sin B

1
cos(A  B)
 tan A tan B  tan 2 C (proved) ]
2 . cos A  cos B  cos 3 B and
*If
sin A cos(A  B)
sin A
1
2 . sin A  sin B  sin 3 B ,P.T, sin( A  B)   .
3
[HINTS: 2.cos A  cos B  cos3 B  (i) and 2.sin A  sin B  sin 3 B (ii)
Applying (ii)×cosB – (i) ×sinB,
2.sin(A  B)  sin B.cos B  sin 3 B.cos B  sin B.cos B  cos3 B.sin B
1
  sin B.cos B   sin 2B  (iii)
2
Adding the squares of (i) & (ii),

2  cos B  cos3 B
2
 
 sin B  sin 3 B

2
 1  2(cos 4 B  sin 4 B)  cos6 B  sin 6 B  2
 1  2cos 2B  1  3cos 2 B.sin 2 B  2
 8cos 2B  3  4cos 2 B.sin 2 B  0
 8cos 2B  3  sin 2 2B  0
 8cos 2B  3  3cos 2 2B  0
 3cos 2 2B  8cos 2B  3  0
1
sin ce cos 2B  3
3
1 1 1
1
From (iii)
sin 2 (A  B)   1   
 sin(A  B)   proved.]
8 9 9
3
Find a & b such that the inequality a ≤ 3.cos x + 5.sin(x– /6) ≤ b holds good  x.

5 3
5
1
5 3

[HINTS: 3cos x  5sin  x    3cos x 
sin x  cos x  cos x 
sin x
6
2
2
2
2

  3cos 2B  1 cos 2B  3  0
5.
 cos 2B  
 1

5 3
1
5 3
 19 
cos x 
sin x   19.cos  x    [where, cos  
& sin  
]
2
19
2
19
2
19
2
19


Now
x,
 19  19.cos  x     19
Hence a   19
 (0342) 262 4499
& b  19
CONFI-HS-PR&SOL-TRIG-02-161304
A Mathematics Center for Excellence