Trigonometric Functions Practice Questions - DPP for JEE

 Page 2


?
?
?
As  (cos x – sec x)
2
 = 0  or  positive
y = 2  or  y  2
2. (c)
Hence period = 
3. (c) Given an angle ?  which is divided into two parts A and B such
that A – B = k and A + B = ,
and tan A : tan B = k : 1, i.e. 
?
(by componendo and dividendo)
 sin ?
4. (d)   The given equation can be written as
? 
and 
5. (d) Consider tan  (1 + sec?) = tan 
Page 3


?
?
?
As  (cos x – sec x)
2
 = 0  or  positive
y = 2  or  y  2
2. (c)
Hence period = 
3. (c) Given an angle ?  which is divided into two parts A and B such
that A – B = k and A + B = ,
and tan A : tan B = k : 1, i.e. 
?
(by componendo and dividendo)
 sin ?
4. (d)   The given equation can be written as
? 
and 
5. (d) Consider tan  (1 + sec?) = tan 
=  … =       …(1)
? f
1
(?) = tan?/2 (1 + sec?) (1 + sec 2?)
= [tan ?/2 (1 + sec ?)] (1 + sec 2?)
= (tan ?) (1 + sec 2?) [from(1)]
= tan 2? [replacing ? by 2? as above]
? f
1
(?) = tan 2
1
 ? …(2)
Similarly, f
2
(?) = tan 2
2 
?, f
3
(?)
= tan 2
3
?, f
4
(?) = tan2
4
?   etc.
?
6. (b) The given expression can be written as
7. (a) We have sin a + sin ß + sin ? – sin ( a + ß + ?)
= sin a + sin ß + sin ? – sin a cos ß cos ?
– cos a sin ß cos ? – cos a cos ß sin ?
+ sin a sin ß sin ?
= sin a (1 – cos ß cos ?) + sin ß (1 – cos a cos ?)
+ sin ?(1–cos a cosß) + sin a sin ß sin ? > 0
? sin a + sin ß + sin ? > sin (a + ß + ?)
Trick :Put a = 30, ß = 30, ? = 60 and check...
?
Page 4


?
?
?
As  (cos x – sec x)
2
 = 0  or  positive
y = 2  or  y  2
2. (c)
Hence period = 
3. (c) Given an angle ?  which is divided into two parts A and B such
that A – B = k and A + B = ,
and tan A : tan B = k : 1, i.e. 
?
(by componendo and dividendo)
 sin ?
4. (d)   The given equation can be written as
? 
and 
5. (d) Consider tan  (1 + sec?) = tan 
=  … =       …(1)
? f
1
(?) = tan?/2 (1 + sec?) (1 + sec 2?)
= [tan ?/2 (1 + sec ?)] (1 + sec 2?)
= (tan ?) (1 + sec 2?) [from(1)]
= tan 2? [replacing ? by 2? as above]
? f
1
(?) = tan 2
1
 ? …(2)
Similarly, f
2
(?) = tan 2
2 
?, f
3
(?)
= tan 2
3
?, f
4
(?) = tan2
4
?   etc.
?
6. (b) The given expression can be written as
7. (a) We have sin a + sin ß + sin ? – sin ( a + ß + ?)
= sin a + sin ß + sin ? – sin a cos ß cos ?
– cos a sin ß cos ? – cos a cos ß sin ?
+ sin a sin ß sin ?
= sin a (1 – cos ß cos ?) + sin ß (1 – cos a cos ?)
+ sin ?(1–cos a cosß) + sin a sin ß sin ? > 0
? sin a + sin ß + sin ? > sin (a + ß + ?)
Trick :Put a = 30, ß = 30, ? = 60 and check...
?
8. (b) Consider 
9. (a) Let  cos a, and  sin a
? i.e. a = p/12
From the equation, r cos (? – a) = 2
? cos (? – p /12)  = cos (p/4)
? ? = 2np ± p/4 + p/12
10. (a) We have, 
? 
? Either 
or  where k is an odd integer
? 
For least positive non-integral solution
is 
11. (b) Given, 3 cos
2
A +2 cos
2
 B = 4
 = 3 sin
2
A ...(1)
and 2 cos B sin B = 3 sin A cos A
sin 2B = 3 sin A cos A ...(2)
Now,  cos (A + 2B) = cos A cos 2B – sin A sin 2B
= cos A (3 sin
2
 A) – sin A (3 sin A cos A) = 0
[using eqs. (1) and (2)]
Page 5


?
?
?
As  (cos x – sec x)
2
 = 0  or  positive
y = 2  or  y  2
2. (c)
Hence period = 
3. (c) Given an angle ?  which is divided into two parts A and B such
that A – B = k and A + B = ,
and tan A : tan B = k : 1, i.e. 
?
(by componendo and dividendo)
 sin ?
4. (d)   The given equation can be written as
? 
and 
5. (d) Consider tan  (1 + sec?) = tan 
=  … =       …(1)
? f
1
(?) = tan?/2 (1 + sec?) (1 + sec 2?)
= [tan ?/2 (1 + sec ?)] (1 + sec 2?)
= (tan ?) (1 + sec 2?) [from(1)]
= tan 2? [replacing ? by 2? as above]
? f
1
(?) = tan 2
1
 ? …(2)
Similarly, f
2
(?) = tan 2
2 
?, f
3
(?)
= tan 2
3
?, f
4
(?) = tan2
4
?   etc.
?
6. (b) The given expression can be written as
7. (a) We have sin a + sin ß + sin ? – sin ( a + ß + ?)
= sin a + sin ß + sin ? – sin a cos ß cos ?
– cos a sin ß cos ? – cos a cos ß sin ?
+ sin a sin ß sin ?
= sin a (1 – cos ß cos ?) + sin ß (1 – cos a cos ?)
+ sin ?(1–cos a cosß) + sin a sin ß sin ? > 0
? sin a + sin ß + sin ? > sin (a + ß + ?)
Trick :Put a = 30, ß = 30, ? = 60 and check...
?
8. (b) Consider 
9. (a) Let  cos a, and  sin a
? i.e. a = p/12
From the equation, r cos (? – a) = 2
? cos (? – p /12)  = cos (p/4)
? ? = 2np ± p/4 + p/12
10. (a) We have, 
? 
? Either 
or  where k is an odd integer
? 
For least positive non-integral solution
is 
11. (b) Given, 3 cos
2
A +2 cos
2
 B = 4
 = 3 sin
2
A ...(1)
and 2 cos B sin B = 3 sin A cos A
sin 2B = 3 sin A cos A ...(2)
Now,  cos (A + 2B) = cos A cos 2B – sin A sin 2B
= cos A (3 sin
2
 A) – sin A (3 sin A cos A) = 0
[using eqs. (1) and (2)]
12. (c) tan (cot x) = cot (tan x) = tan 
? cot x = n – tan x 
[  tan ?  = tan a ? ? = np + a]
? cot x + tan x = n 
? = (2n + 1) 
? = (2n + 1) 
?
? sin 2x = 
13. (a) Given f (x) = cos (log x)
?  f (xy) = cos (log xy)
f (xy) = cos [log x + log y] ....(i)
And  = cos
 = cos (log x – log y) ....(ii)
Adding (i) and (ii), we get
f (xy) +  = cos (log x + log y) + cos (log x – logy)
= 2 cos (log x). cos (log y)
?  f (xy) +  = 2 f (x). f (y)