trignometry basic questions

L EC TURE

22 Trigonometric Functions

1. TABLE REPRESENTING ALL TRIGONAL RATIOS (TRs) IN TERMS OF ONE TRIGONAL RATIO

sinθ cosθ tanθ cotθ secθ cosecθ

sinθ sinθ 21 cos− θ 2

tan1 tan

θ

+ θ 2

11 cot+ θ

2sec 1sec

θ−θ

1cosec θ

cosθ 21 sin− θ cosθ

2

11 tan+ θ 2

cot1 cot

θ

+ θ

1secθ

2cosec 1cosec

θ−θ

tanθ 2

sin1 sin

θ

− θ

21 coscos− θ

θtanθ

1cotθ 2sec 1θ− 2

1cosec 1θ−

cotθ21 sin

sin− θ

θ 2

cos1 cos

θ

− θ

1tanθ cotθ 2

1sec 1θ−

2cosec 1θ−

secθ 2

11 sin− θ

1cosθ 21 tan+ θ

21 cotcot+ θ

θsecθ 2

coseccosec 1

θ

−

1. TABLE REPRESENTING ALL TRIGONAL RATIOS (TR

2. T-RATIOS (OR TRIGONOMETRIC FUNCTIONS)

sin ,cos , tan ,p b p

h h bθ = θ = θ =

cosec ,sec ,cot ;h h b

p b pθ = θ = θ =

2. T-RATIOS (OR TRIGONOMETRIC

O

p

B

θA

h

bp + b = h2 2 2

‘p’ perpendicular’, ‘b’ base and ‘h’ stands for hypotenuse.

A.18 Trigonometric Functions

3. FOLLOWING ARE SOME OF THE FUNDAMENTAL TRIGONOMETRIC IDENTITIES

(i) 1sincosec

θ =θ

or 1cosecsin

θ =θ

(ii) 1cossec

θ =θ

or 1seccos

θ =θ

(iii) 1cottan

θ =θ

or 1tancot

θ =θ

(iv) sintancos

θθ =

θ

or coscotsin

θθ =

θ (v) sin2θ+cos2θ=1orsin2 θ=1–cos2θorcos2

θ=1–sin2 θ (vi) sec2θ–tan2θ=1 or1+tan2θ–sin2θ

or 1sec tansec tan

θ− θ =θ+ θ

(vii) cosec2θ–cot2θ=1or1+cot2θ=cosec2θ

or 1cosec cotcosec cot

θ− θ =θ+ θ

4. MAXIMUM AND MINIMUM VALUES OF TRIGONOMETRICAL FUNCTIONS

(i) –1≤ sin x ≤ 1 (ii) –1≤ cos x ≤ 1 (iii) –∞ < tan x < ∞ (iv) –∞ < cot x < ∞ (v) |sec x | ≥ 1 i.e., sec x ≤–1orsecx ≥ 1 (vi) |cosec x| ≥ 1 i.e., cosec x ≤–1orcosecx ≥ 1

3. FOLLOWING ARE SOME OF THE

4. MAXIMUM AND MINIMUM VALUES OF

(vii) sin2 x +cosec2 x ≥ 2, ∀ x ∈ R (viii) cos2 x +sec2 x ≥ 2, ∀ x ∈ R (ix) tan2 x +cot2 x ≥ 2, ∀ x ∈ R (x) |sin x+cosecx | ≥ 2 (xi) |cos x+secx | ≥ 2 (xii) |tan x+cotx | ≥ 2 (xiii) If a sin x+b cos x=c, then (a cos x–b sin x)2

=a2 + b2 – c2

5. SOME USEFUL RESULTS

(i) sin2θ+cos4θ=sin4θ+cos2θ=1–sin2θcos2θ (ii) sin4θ+cos4θ=1–2sin2θcos2θ (iii) sin6θ+cos6 θ=1–3sin2θcos2θ (iv) sin4θ–cos4 θ=1–2cos2θ (v) sin8θ–cos8θ=(sin2θ–cos2θ)(1–2sin2θ

cos2θ) (vi) sec2θ+cosec2 θ=sec2θ–cosec2θ=tan2θ+

cot2θ+2

(vii) 1 1 sinsec tansec tan 1 sin

+ θ= θ+ θ =

θ− θ − θ

(viii) 1 1 sinsec tansec tan 1 sin

− θ= θ− θ =

θ+ θ + θ

(ix) 1 1 coscosec cotcosec cot 1 cos

+ θ= θ+ θ =

θ− θ − θ

(x) 1 1 coscosec cotcosec cot 1 cos

− θ= θ− θ =

θ+ θ + θ

(xi) 1 sin cos sec tancos 1 sin− θ θ

= = θ− θθ + θ

(xii) 1 cos sin cosec cotsin 1 cos− θ θ

= = θ− θθ + θ

(xiii) 1 sin cos sec tancos 1 sin− θ θ

= = θ− θθ + θ

(xiv) 1 cos sin cosec cotsin 1 cos− θ θ

= = θ− θθ + θ

5. SOME USEFUL RESULTS

(i) sin

Trigonometric Functions A.19

1. Show that 3 3

(1 tan cot )(sin cos )sec cosec

+ θ+ θ θ− θθ− θ

Solution

L.H.S.= 3 3

3 3

sin cos1 (sin cos )(1 tan cot )(sin cos ) cos sin

1 1sec coseccos sin

θ θ + + θ− θ + θ+ θ θ− θ θ θ =θ− θ −

θ θ

3 3

3 3

sin cos1 (sin cos )(1 tan cot )(sin cos ) cos sin

1 1sec coseccos sin

θ θ + + θ− θ + θ+ θ θ− θ θ θ =θ− θ −

θ θ

2 2

3 3

3 3

(sin cos sin cos )(sin cos )sin cossin cossin cos

θ θ+ θ+ θ θ− θ=

θ− θθ θ θ θ

3 3 3 3

3 3

(sin cos ) sin cossin cos (sin cos )

θ− θ θ θ=

θ θ θ− θ

2 2 3 3( ( )( ) )a b ab a b a b+ + − = −

2 2sin cos= θ θ = R.H.S.

2. If cos2α–sin2α=tan2β,thenshowthattan2 α=cos2β–sin2β.

Solution

Given, cos2α–sin2α=tan2β2 2 2

2 2 2

cos sin sincos sin cos

α − α β⇒ =

α + α β

2 2 2

2 2 2

cos sin coscos sin sin

α + α β⇒ =

α − α β

2 2( cos sin 1)∴ α+ α =Applying componendo and dividendo, we get

2 2 2 2 2 2

2 2 2 2 2 2

cos sin cos sin cos sincos sin cos sin cos sin

α + α + α − α β+ β=

α + α − α + α β− β2

2 2 2

2cos 12sin cos sin

α⇒ =

α β− β2

2 22

sin cos sincos

α⇒ = β− β

α2 2 2tan cos sin ,⇒ α = β− β as desired.

3. If sinsin

A mB= and cos

cosA nB= ,findthevalueof

tan B; n2 < 1 < m2.

Solution

Given sin sin sinsin

A m A m BB= ⇒ = .........(1)

and cos cos coscos

A n A n BB= ⇒ = ..........(2)

By squaring (1) and (2) and adding, we get

2 2 2 21 sin cosm B n B= +2

2 22 2

1 sincos cos

Bm nB B

⇒ = + (Dividing by cos2 B)

2 2 2 2sec tan⇒ = +B m B n

2 2 2 21 tan tan⇒ + = +B m B n 2

2 2 22

11 ( 1) tan tan1

nn m B Bm

2 −⇒ − = − ⇒ =

−

2

2

1tan1

nBm−

⇒ = ±−

4. Eliminateθbetween cosec sin ,θ− θ = a sec cosθ− θ = b.

Solution

Given cosec sin aθ− θ = .............(1)and sec cos bθ− θ = .............(2)

From (1), 21 cossin

sin sina aθ

− θ = ⇒ =θ θ

.............(3)

From (2), 21 sincos

cos cosb bθ

− θ = ⇒ =θ θ

..............(4)Squaring(3)andmultiplyingby(4),weget

3 2 2 1/3cos cos ( )a b a bθ = ⇒ θ = ............(5)

Squaring(4)andmultiplyingby(3),weget3 2 2 1/3sin sin ( )ab abθ = ⇒ θ = ............(6)


5. Eliminateθandφ between cos cos , cos sin , sin .x r y r z r= θ φ = θ φ = θ

Solution

Given cos cos , cos sinx r y r= θ φ = θ φ andsin .= θz r

Squaring and adding, we get2 2 2+ +x y z

2 2 2 2 2 2 2 2cos cos cos sin sinr r r= θ φ+ θ φ+ θ2 2 2 2 2 2cos (cos sin ) sinr r= θ φ+ φ + θ2 2 2 2(sin cos )r r= θ+ θ =

2 2[ sin cos 1]∴ θ+ θ =

6. If sin cosandsin cos

m nα α= =

β β, than prove that

22 2

2

1tan ; 1 .1

m n n mn m

−α = ± < <

−

Solution

Given sinsin

mα=

β ............(1)

and coscos

nα=

β .............(2)

From (1), sinsinmα

β = ............(3)

and from (2), coscosnα

β = .............(4)

Squaring(3)and(4)andadding,weget

2 2

2 2

sin cos1m nα α

= +

2 2 2 2 2 2sin cosm n n m⇒ = α+ α

Divide throughtout by cos2α2 2 2 2 2 2

2 2 2

sin coscos cos cosm n n mα α

⇒ = +α α α

2 2 2 2 2 2sec tanm n n m⇒ α = α +2 2 2 2 2 2(1 tan ) tanm n n m⇒ + α = α +2 2 2 2 2 2 2 2tan tanm n m n n m⇒ + α = α +2 2 2 2 2 2 2 2tan tanm n m n m n⇒ − = α − α

2 22

2 2

1tan1

m nn m

−⇒ α = ×

−2

22

1tan .1

m nn m

−⇒ α = ±

− (Proved)

7. If 4 4sin cos 1A A

a b a b+ =

+, then prove that

8 8

3 3 3

sin cos 1( )

A Aa b a b

+ =+

Solution

(a+b) (b sin4 A+a cos 4A)–ab=0

or ab [sin4 A+cos 4A–1]+a2 cos4A+b2 sin4A=0

or ab[1–2sin2A cos2A–1]+a2 cos4A+b2 sin4A=0

or (a cos2 A–b sin 2 A)2=0

or 2 2cos sin 1A A

b a a b= =

+

8 8 4 4

3 3 3 4 3 4

sin cos 1 1. .( ) ( )

A A a ba b a a b b a b

∴ + = ++ +

4 3

( ) 1( ) ( )

a ba b a b+

= =+ +

8. If sin x+sin2 x+sin3 x=1,then

cos6 x–4cos4 x+8cos2 x=........................

Solution

The given relation can be written as sin x(1+sin2 x)=1–sin2 x=cos2 x⇒ sin x(2–cos 2 x)=cos 2 x⇒ sin2 x(2–cos2 x)2=cos4 x (squaring both sides)⇒ (1–cos2 x)(4–4cos2 x+cos4 x)=cos4 x ⇒ cos6 x–4cos4 x+8cos2 x=4.

9. If 8 sinθ = 4 + cosθ, then find the value ofsinθ.

Solution

8sinθ=4+cosθ⇒ 8sinθ–4=cosθ⇒ (8sinθ–4)2=cos2θ⇒ (8sinθ–4)2=1–sin2θ⇒ 64 sin2θ+16–64sinθ=1–sin2θ⇒ 65 sin2θ–64sinθ+15=0⇒ 65 sin2θ–39sinθ–25sinθ+15=0


⇒ 13sinθ(5sinθ–3)–5(5sinθ–3)=0⇒ (13sinθ–5)(5sinθ–3)=0

⇒ 5 3sin or .13 5

θ =

⇒ 3sin5

θ = infirstquadrant

and 5sin13

θ = in second quadrant

10. Prove that, tan cot sec cosec 1

1 cot 1 tanθ θ

+ = θ θ+− θ − θ

Solution

L.H.S. tan cot1 cot 1 tan

θ θ= +

− θ − θ

sin coscos sin

sin cos cos sinsin cos

θ θθ θ= +

θ− θ θ− θθ θ

2 2sin coscos (sin cos ) sin (cos sin )

θ θ= +

θ θ− θ θ θ− θ

3 3sin coscos sin (sin cos )

θ− θ=

θ θ θ− θ

3 3 2 2[ ( )( )]− = − + +a b a b a b ab

2 2(sin cos ) (sin cos sin cos )

cos sin (sin cos )θ− θ θ+ θ + θ θ

=θ θ θ− θ

1 sin cos sec cosec 1sin cos+ θ θ

= = θ θ+θ θ

EXERCISE 1

1. Prove that, sec4θ–sec2 θ=tan4θ+tan2θ

2. Prove that, sin 1 cos 2cosec 1+cos sin

θ + θ+ = θ

θ θ

3. Prove that, 2cosec cosec 2seccosec 1 cosec 1

θ θ+ = θ

− +

4. Prove that, (cosecθ–sin θ)(secθ–cosθ)(tanθ+cotθ)=1

5. If tan θ = 34, then find all the trigonometric

ratios.

6. Ifsecθ+tanθ=4,findthevaluesofsinθ,cosθ,secθandtanθ.

7. Prove that, 21 cos (cosec cot )1 cos

x x xx

+= +

−

8. If sin x+sin2 x=1,thenprovethatcos 2 x+cos4 x=1

EXERCISE 1

1. Prove that, sec

9. Ifsecθ+tanθ=p,obtainthevaluesofsecθ,tanθintermsofp.

10. Prove that, sin cot sin cosec 1 cosθ θ+ θ θ = + θ

11. Prove that, 2 2

2 2 2 2 2

tan cosec 1tan 1 sec cosec sin cos

θ θ+ =

θ− θ− θ θ− θ

12. Prove that, 2 2 2(1 tan ) (1 cot ) (sec cosec )x x x x− + − = − .

EXERCISE 2

1. Prove that tan2θ–sin2θ=tan2θ–sin2θ

2. Prove that2sec tan 1 2sec tan 2tan

sec tanθ− θ

= − θ θ+ θθ+ θ

3. Prove that 1 cos (cosec cot )1 cos+ θ

= θ+ θ− θ

EXERCISE 2

1. Prove that


4. Prove that 1 cos sinsin 1 cos− θ θ

=θ + θ

5. Prove that cot4θ+cot2θ=cosec4θ–cosec2θ

6. Prove that (Sin A+cosA) (tan A+cotA)=sec A+cosecA

7. Prove that cos cosec sin sec cosec sec

cos sinA A A A A A

A A−

= −+

8. Prove that 2 sin2 A+cos4 A=1+sin4 A

9. Prove that 1 1 1 1

cosec cot sin sin cosec cot− = −

θ− θ θ θ θ+ θ

10. Prove that, 21 cos (cosec cot )1 cos

x x xx

+= +

−

11. If sin x+sin2 x=1,provethatcos2 x +cos4 x =1

12. Ifsecθ+tanθ=p,obtainthevaluesofsecθ,tanθintermsofp.

EXERCISE 1

5. 3 4 4 5 5sin ,cos ,cot ,cosec ,sec5 5 3 3 4

θ = θ = θ = θ = θ =

3 4 4 5 5sin ,cos ,cot ,cosec ,sec5 5 3 3 4

θ = θ = θ = θ = θ =

5cosec ,3

θ =

3 4 4 5 5sin ,cos ,cot ,cosec ,sec5 5 3 3 4

θ = θ = θ = θ = θ =

6. 15 8 15 17sin ,cos , tan ,sec17 17 8 8

θ = θ = θ = θ =

15 8 15 17sin ,cos , tan ,sec17 17 8 8

θ = θ = θ = θ =

15 8 15 17sin ,cos , tan ,sec17 17 8 8

θ = θ = θ = θ =

9. 2 2 2

2

1 1 1sec , tan and sin =2 2 1

p p pp p p+ − −

θ = θ = θ+

2 2 2

2


p p pp p p+ − −

θ = θ = θ+

EXERCISE 2

12. =2 2 2

2


p p pp p p+ − −

θ = θ = θ+

and sin θ 2 2 2

2


p p pp p p+ − −

θ = θ = θ+

1. If 2sin 1 cos sinthen

1 cos sin 1 sinay − α + α

=+ α + α + α

is equal to [UPSEAT, 99](a) 1/y (b) y(c) 1–y (d) 1+y

Solution

(b) 1 cos sin1 sin

− α + α+ α

1 cos sin 1 cos sin.

1 sin 1 cos sin− α + α + α + α

=+ α + α + α

2 2(1 sin ) cos(1 sin )(1 cos sin )

+ α − α=

+ α + α + α

2 2(1 sin 2sin ) (1 sin )

(1 sin )(1 cos sin )+ α + α − − α

=+ α + α + α

2sin (1 sin )(1 sin )(1 cos sin )

α + α=

+ α + α + α

2sin1 cos sin

yα= =

+ α + α


2. If a cos3α+3a cosαsin2α=m and a sin3α+3a cos2αsinα=n, then (m+n)2/3+(m–n)2/3 is equal to(a) 2a2 (b) 2a1/3

(c) 2a2/3 (d) 2a3

Solution

(c) From the given relations, we get m+n=a cos3α+3a cosαsin2 α+3a

cos2 αsinα+a sin3α =a(cosα+sinα)3 similarly, m–n=

a(cosα–sinα)3 ∴ (m+n)2/3+(m–n)2/3=a2/3

[(cosα+sinα)2+(cosα–sinα)2] =a2/3 [2 (cos2 α+sin2 α)]=2a2/3

3. (m+2)sinθ+(2m–1)cosθ=2m+1,if(a) tanθ=3/4(b) tanθ=4/3(c) tanθ=2m/(m2 –1)(d) tanθ=2m/(m2+1)

Solution

(b, c) The given relation can be written as (m+2) tanθ+ (2m –1)= (2m+1)

secθ⇒ (m + 2)2 tan2θ + 2(m + 2) (2m – 1)

tanθ+(2m–1)2 =(2m+1)2(1+tan2θ)⇒ [(m+2)2 –(2m+1)2 ] tan2θ+2(m+2)

(2m–1)tanθ +(2m–1)2–(2m+1)2=0⇒ 3(1–m2) tan2θ+(4m2+6m–4)tanθ

–8m=0

⇒ [(m+2)2 –(2m+1)2 ] tan2θ+2(m+2)(2m–1)tanθ

+(2m–1)2–(2m+1)2=0⇒ 3(1–m2) tan2θ+(4m2+6m–4)tanθ–

8m=0

⇒ (3tanθ–4)[(1–m2)tanθ+2m]=0Which is true if tanθ =4/3 or tanθ = 2m/ (m2–1)

4. If x=secϕ–tanϕ and y=cosecϕ+cotϕ, then

(a) 11

yxy+

=−

(b) 11

yxy−

=+

(c) 11

xyx

+=

−

(d) xy+x–y+1=0

Solution

(b, c, d) We have 1 sin 1 cos,cos sin

x y− φ + φ= =

φ φ

By multiplying, we get

(1 sin )(1 cos )cos sin

xy − φ + φ=

φ φ

⇒ xy+1

1 sin cos sin cos sin coscos sin

− φ+ φ− φ φ+ φ φ=

φ φ

1 sin cos

cos sin− φ+ φ

=φ φ

and

(1 sin )sin cos (1 cos )cos sin

− φ φ− φ + φ− =

φ φx y

2 2sin sin cos cos

cos sinφ− φ− φ− φ

=φ φ

sin cos 1 ( 1)

cos sinφ− φ−

= = − +φ φ

xy

Thus, 11 01

yxy x y xy−

= − + = ⇒ =+

and

11

xyx

+=

−

5. If 21 sin 3sin cos ,A A A+ = then possible values of tan A are [NDA-2005](a) 1,1/2 (b) 2,1/4(c) 3,1/6 (d) 4,1/8

Solution

(a) 21 sin 3sin cosA A A+ =2 2 2(sin cos ) sin 3sin cosA A A A A⇒ + + =2 2

2

2sin 3sin cos cos 02tan 3tan 1 0

A A A AA A

⇒ − + =

⇒ − + =

A.24 Trigonometric Functions2 2

2

2sin 3sin cos cos 02tan 3tan 1 0

A A A AA A

⇒ − + =

⇒ − + =

(tan 1)(2 tan 1) 0⇒ − − =A Atan 1 0,2 tan 1 0A A⇒ − = − =

⇒tan A–1=0,2tanA–1=0⇒tan A=1,1/2

6. For what values of x is theequation2sinθ= x+1/x is valid? [NDA-2006](a) x=±1(b) all real values of x(c) –1<x < 1(d) x > 1 and x<–1

Solution

(a) x2 – 2xsinθ+1=0

22sin 4sin 42

θ± θ−⇒ =x

2sin sin 1= θ± θ−

Whichisvalidwhensinθ=±1Thenx=±1.

7. If x = r sinθ cosϕ, y = r sinθ sinϕ and z = r cosθ, then x2 +y2 + z2 is independent of which of the following? [NDA-2007](a) r only (b) r, ϕ(c) θ,ϕ (d) r,θ

Solution

(c) x2+y2+z2 =r2 sin2θcos2ϕ+r2 sin2θsin2ϕ+r2 cos2θ=r2 sin2θ(cos2ϕ+sin2ϕ)+ r2 cos2θ=r2 (sin2θ+cos2θ)=r2.

8. What is the value of (secθ–cosθ)(cosecθ–sinθ)(cotθ+tanθ)? [NDA-07](a) 1 (b) 2(c) sinθ (d) cosθ

Solution

(a) (secθ– cosθ) (cosecθ– sinθ) (cotθ+tanθ)

2 2 2(1 cos ) (1 sin ) (1 tan )

cos sin tan− θ − θ + θ

= × ×θ θ θ

2 2 2sin .cos .sec .cos 1.

cos .sin .sinθ θ θ θ

= =θ θ θ

9. If sin x+sin2 x=1,thencos6 x+cos 12 x+3cos10x+3cos8 x is equal to [Pb. CET-2002; MPPET-2006](a) 1 (b) cos3 x sin3 x(c) 0 (d)∞

Solution

(a) ∵ sin x+sin2 x=1⇒ sin x=1–sin2 x ⇒ sin x=cos2 x∴ cos6 x+cos12 x+3cos10 x+3cos8 x =sin3 x+sin6 x+3sin5 x+3sin4 x=

(sin x+sin2 x)3=1.

10. The value of the expression [Bihar EE-1990]

2sin 1 cos sin11 cos sin 1 cos

y y yy y y

+− + −

+ − is equal to

(a) 0 (b) 1(c) sin y (d) cos y

Solution

(d) The given value

= 1 – (1 – cos y) +2 21 cos sin

sin (1 cos )y y

y y− −

− =

cos y+0=cosy

11. 2

2

2sin tan (1 tan ) 2sin sec(1 tan )

θ θ − θ + θ θ+ θ

is equal

to [Roorkee-1975]

(a) sin1 tan

θ+ θ

(b) 2sin1 tan

θ+ θ

(c) 2

2sin(1 tan )

θ+ θ

(d) None of these

Solution

(b) Given that expression

22

2sin {tan (1 tan ) sec }(1 tan )

θ= θ − θ + θ

+ θ

2 22

2sin {tan tan 1 tan }(1 tan )

θ= θ− θ+ + θ

+ θ

2sin1 tan

θ=

+ θ


12. Ifcotθ+tanθ=mandsecθ–cosθ=n, then which of the following is correct? (a) m(mn2)1/3–n (nm2)1/3=1(b) m(m2n)1/3–n (mn2)1/3=1(c) n(mn2)1/3–m (nm2)1/3=1(d) n(m2n)1/3–m (mn2) 1/3=1

Solution

As given, 21 tan 1 tan tan

tanm m+ θ = ⇒ + θ = θ

θ

⇒ sec2θ=mtanθ .................(i)and secθ–cosθ=n⇒ sec2θ–1=nsecθ⇒ tan2θ=nsecθ⇒ tan4θ=n2 sec2 θ=n2 mtanθ{by(i)}⇒ tan3θ=n2m, [∵tanθ≠0)⇒ tanθ=(n2m)1/3 ................ (ii) Also, sec2θ=mtanθ=m(n2m)1/3

{by (i) and (ii)}using the identity sec2θ–tan2θ=1⇒ m(n2m)1/3–(n2 m)2/3=1⇒ m(mn2)1/3–n (nm2)1/3=1

13. 22

4sin( )

xyx y

θ =+

is true, if and only if [AIEEE-2002](a) x–y ≠0 (b) x=–y(c) x=y (d) x ≠0,y ≠0

Solution

(c) Since,θ≤1

⇒

2

4 1( )

xyx y

⇒ ≤+

22

4sin given( )

xyx y

θ = +

⇒ x2+y2+2xy–4xy ≥0

⇒ (x–y)2 ≥0

Which is true for all real values of x and y provided x+y ≠0,otherwise, 2

4( )

xyx y+

will be meaningless.

14. If y=sin2θ+cosec2θ,θ≠0,then [AIEEE-2002](a) y=0 (b) y≤0(c) y ≥–2 (d) y ≥ 2

Solution

(d) Given that, y=sin2θ+cosec2θ∴ y=(sinθ–cosecθ)2+2⇒ y ≥2,θ≠0

15. If cos x=tany, cos y=tanz, cos z=tanx; than prove that sin x=siny=sinz=2sin18o.

Solution

Making use of given relations, we have cos2 x=tan 2 y=sec2 y–1=cot2 z–1.

or 2 2

22 2

cos tan1 cos1 cos 1 tan

z xxz x

+ = =− −

or 2

22 2

sin1 coscos sin

xxx x

+ =−

or changingtosinx,weget(2–sin2 x)(1–2 sin2 x)=sin2x

or 2 sin4 x–6sin2 x+2=0orsin4 x–3sin2

x+1=0

2 3 9 4 3 5 3 5sin2 2 2

x ± − ± −∴ = = =

We have rejected the value 3 52+ as

it is > 1.

or 2

2 6 2 5 5 1sin4 2

x − −

= =

5 1 ( 5 1)sin 2. 2sin182 4− −

∴ = = = °x

By symmetry, we can say that sin x=siny=sin z=2sin18o.

16. Show that, cos (sin ) sin (cos ), 02π

θ > θ ≤ θ ≤ .

[IIT-JEE-1981]

Solution

sin cos 22π

θ+ θ ≤ <

sin cos2π

⇒ θ < − θ

cos (sin ) cos cos sin (cos )2π ⇒ θ > − θ = θ


1. Ifsinθ= 2

21

tt+,thencosθisequalto

(a) 2

21

tt−

(b) 2

21

tt+

(c) 2

2

11

tt

−+

(d) 2

2

11

tt

+−

2. sin cos1 cot 1 tan

θ θ+

− θ − θ is equal to

[Karnataka CET-1998]

(a) 0 (b) 1(c) cosθ–sinθ (d) cosθ+sinθ

3. If for real values of x,cosθ=x+ 1x

, then

[MPPET-1996](a) θisanacuteangle(b) θisarightangle(c) θisanobtuseangle(d) Novaluesofθispossible

4. The equation sec2 θ = 2

4( )

xyx y+

is only

possible when [MPPET-1986; IIT-1996](a) x=y (b) x < y(c) x > y (d) None of these

5. Which of the following relations is correct? [WBJEE-91](a) sin 1< sin 1o

(b) sin 1 > sin 1o

(c) sin1=sin1o

(d) sin1 sin1180π

= °

6. Ifsinθ+cosecθ=2,thensin2θ+cosec2θis equal to [MPPET-1992; MNR-1990; UPSEAT-2002](a) 1 (b) 4(c) 2 (d) None of these

7. Iftanθ= 2021

,cosθwillbe [MPPET-1994]

(a) 2041

± (b) 121

±

(c) 2129

± (d) 2021

±

8. If x=a cos3θ,y=b sin3θ,then(a) (a/x)2/3+(b/y)2/3=1(b) (b/x)2/3+(a/y)2/3=1(c) (x/a)2/3+(y/b)2/3=1(d) (x/b)2/3+(y/a)2/3=1

9. If x=secθ+tanθ,thenx+ 1x

is equal to

[MPPET-1986](a) 1 (b) 2secθ(c) 2 (d) 2tanθ

10. If(1+sinA)(1+sinB)(1+sinC)=(1–sinA) (1–sinB)(1–sinC), then each side is equal to(a) ± sin A sin B sin C(b) ± cosA cosB cos C(c) ± sin A cos B cos C(d) ± cos A sin B sin C

11. If sinθ1+ sinθ2+ sinθ3=3, thencosθ1+cosθ2+cosθ3 is equal to [EAMCET-1994](a) 3 (b) 2(c) 1 (d) 0

12. Iftanθ–cotθ=aandsinθ+cosθ=b, then (b2–1) 2 (a2+4)isequalto [WB JEE-1979](a) 2 (b) –4(c) ± 4 (d) 4

13. Iftanθ=sin

1 cosx

xφ

− φ and tan φ =

sin1 cos

yy

θ− θ

,

then xy

is equal to [MPPET-1991]

(a) sinsin

φθ

(b) sinsin

θφ

(c) sin1 cos

φ− θ

(d) sin1 cos

θ− φ


14. Ifsecθ+tanθ=p,thentanθisequalto [MPPET-1994]

(a) 2

21

pp −

(b) 2 12

pp−

(c) 2 12

pp+ (d) 2

21

pp +

15. If 2sin cosand =

1 cos sin 1 sinp qθ θ=

+ θ+ θ + θ , then

[MPPET-2001]

(a) pq=1 (b) 1qp=

(c) q–p=1 (d) q+p=1

16. Theminimumvalueof9tan2θ+4cot2 θis(a) 13 (b) 9(c) 6 (d) 12

17. The maximum value of 4 sin2 x+3cos2 x is [Karnataka CET-2003](a) 3 (b) 4(c) 5 (d) 7

18. Which value of k, (cosx+sinx) 2 +k sin x cos x–1=0isidentity? [Kerala (Engg.)-2001](a) –1 (b) –2(c) 0 (d) 1

20. If sinx+sin2x=1,thencos8x+2cos6 x+cos4x is equal to(a) 0 (b) –1(c) 2 (d) 1

21. The least value of 2 sin2 θ + 3 cos2 θ is [MPPET-2010]

(a) 1 (b) 2 (c)3 (d)5

22. If 1sec ,4

A xx

= + then the value of sec A+tan A is [MPPET-2010]

(a) 3x (b) 3x

(c)

2x (d) 2x

1. (c) Given 2

2sin1

tt

θ =+

By Pythagoras theoremAC2=AB2+BC2

(1+t2)2=(2t)2+BC2

(1+t2)2=4t2+BC2

⇒ BC2=(1+t2)2–4t2=1+t4+2t2–4t2=1+t4–2t2

⇒ BC2=(1–t2) ⇒ BC=1–t2

∴ 2

2

1cos1

BC tAC t

−θ = =

+

Second Method

22

2 tan2 2sin .1 1 tan

2

tt

θ

θ = =θ+ +

If tan2

t θ=

∴

22

22

1 tan 12cos11 tan

2

tt

θ− −

θ = =θ ++

2. (d) Given sin cos1 cot 1 tan

θ θ+

− θ − θ

sin coscos sin1 1sin cos

θ θ= +

θ θ− −

θ θ

sin cossin cos cos sin

sin cos

θ θ= +

θ− θ θ− θθ θ

⇒ 2 2sin cos

sin cos cos sinθ θ

+θ− θ θ− θ


⇒ 2 2sin cos

sin cos sin cosθ θ

−θ− θ θ− θ

⇒ 2 2sin cos

sin cosθ− θθ− θ

⇒ (sin cos )(sin cos )(sin cos )

θ+ θ θ− θθ− θ

⇒ sinθ+cosθ

3. (d)Givenequationiscosθ=x+1/xor x2–xcosθ+1=0for real value of x By B2 > 4AC

∵ A=1,B=–cosθ,C=1

cos2θ>4(1)(1)or|cosθ|≤2

Whichisimpossiblebecause|cosθ|<1∴ Norealvalueofθispossible.

4. (a) sec2θ≥1∀θ⇒ 2

4 1( )

xyx y

≥+

⇒ 4xy≥(x + y)2

⇒ (x + y)2–4xy≤0

⇒ (x – y)2≤0.But(x – y)2≥0

⇒ (x – y)2=0⇒ x=y

5. (a) 1 radian 180=

πdegrees=57°(approxi-

mately)⇒ sinx is increasing function

1° < 1c , sin (1°) < sin (1c)

6. (c)sinθ+cosec(θ)=2

⇒ 1sin 2sin

θ+ =θ

⇒ sin2θ–2sinθ+1=0

⇒ (sinθ–1)2=0

⇒ sinθ=1

⇒ θ=π/2∴ sin2(θ)+cosec2(θ)=1+1=2

7. (c) Given 20tan21

θ = Squaring both sides

2 400tan441

θ =

By sec2θ=1+tan2θ

∴ 2 400sec 1441

θ = +

or 2 441cos841

θ = or 21cos29

θ = ±

8. (c) Given x=a cos3θ,y=b sin3θ

⇒ 3 3cos , sinx ya b= θ = θ

taking cube root on both sides.

⇒ 1/3 1/3

cos , sinx ya b

= θ = θ

Now square and add2 21 1

3 3x ya b

+

=sin2θ+cos2θ

⇒ 2/3 2/3

1x ya b

+ =

9. (b) 1xx

+ = (secθ+tanθ)+ 1sec tanθ+ θ

=(secθ+tanθ)+ 2 2

(sec tan )sec tan

θ− θθ− θ

⇒ 1 2 sec( )xx

+ = θ

10. (b) Multiplying both sides by (1 – sin A) (1–sin2 B)(1–sin2 C)Wehave (1–sin2 A)(1–sin2 B)(1–sin2 C) =(1–sinA)2(1–sinB)2(1–sinC)2

⇒ (1–sinA)(1–sinB)(1–sinC)=±cosA cos B cos C

Similarly,(1+sinA)(1+sinB)(1+sinC) =±cosA cos B cos C

11. (d)Givensinθ1+sinθ2+sinθ3=3

⇒ sinθ1=1=sinθ2=sinθ3⇒ θ1=θ2=θ3=π/2

⇒ cosθ1=cosθ2=cosθ3=0⇒ cosθ1+cosθ2+cosθ3=0

12. (d)Given,tanθ–cotθ=a …….. (1)sinθ+cosθ=b ……….. (2)


(2) ⇒ b2=1+2sinθcosθ⇒ b2–1=sin(2θ) .................(3)(1) ⇒ a2=tan2θ+cot2θ–2tanθcotθ=tan2θ+cot2θ–2=a2+4=(tanθ+cotθ)2

=a2+42sin cos

cos sinθ θ = + θ θ

22 2sin cos 2sin cos sin 2

θ+ θ = = θ θ θ

2

4( 1)b

=−

⇒ (a2+4)(b2–1)2=4

13. (b) sintan1 cos

xx

φθ =

− φ

⇒ (1–x cos ϕ)tanθ=x sin ϕ

⇒ tansin cos tan

x θ=

φ+ φ φ ………. (1)

Similarly, sintan1 sin

yy

θφ =

− θ

⇒ tansin cos tan

y φ=

θ+ θ θ ………. (2)

(1)(2)

xy

⇒tan sin cos tan.tan sin cos tan

θ θ+ θ φ= φ φ+ φ φ

sin cot cossin cot cos

θ φ+ θ=

φ θ+ φsin (cot cot ) sinsin (cot cot ) sin

θ φ+ θ θ= =

φ θ+ φ φ

14. (b)secθ+tanθ=p …………..(1)

⇒ 1sec tanp

θ− θ = ………… (2)

(1)–(2)⇒ 1 2 tanpp

− = θ ⇒ 2 1tan2

pp−

θ =

15. (d) 2sin cos1 sin cos 1 sin

p q θ θ+ = +

+ θ+ θ + θ

2 22sin 2sin cos sin cos cos(1 sin cos )(1 sin )

θ+ θ+ θ+ θ θ+ θ=

+ θ+ θ + θ2(1 sin cos ) (sin sin sin cos )

(1 sin cos )(1 sin )+ θ+ θ + θ+ θ+ θ θ

=+ θ+ θ + θ

(1 sin cos )(1 sin ) 1(1 sin cos )(1 sin )+ θ+ θ + θ

= =+ θ+ θ + θ

16. (d)Since,A.M.≥G.M.Therefore,

2 22 29tan 4cot 9 tan 4 cot

2θ+ θ

≥ θ× θ

⇒ 2 29tan 4cot 2 36 12θ+ θ ≥ × =

17. (b) f(x)=4sin2 x+3cos2 x=sin2 x+3and 0≤|sinx|≤1

∴ maximum value of sin2 x+3is4.

18. (b) Given(cos x+sinx)2+k sin x cos x–1=0∀x ∈R⇒ cos2 x+sin2 x+2sinx cos x+k sin x cos

x–1=0∀x⇒ (k+2)sinx cos x=0∀x⇒ k+2=0⇒ k=–2

19. (c) x sin3α+y cos3α=sinα.cosα …(1)xsinα–ycosα=0 ……..(2)Now, a2=tan2θ+cot2θ–2tanθ.cotθ=tan2 θ+cot2θ–2⇒ a2+4=(tanθ+cotθ)2

⇒ 2

2 sin cos4cos sin

a θ θ + = + θ θ

22 2sin cos 2sin cos sin 2

θ+ θ = = θ θ θ

2

4( 1)b

=−

⇒ (a2+4)(b2–1)2=4

20. (d) We have,sin x+sin2 x=1⇒ sin x=1–sin2 x⇒ sin x=cos2 x⇒ cos8 x+2cos6 x+cos4 x⇒ sin4 x + 2 sin3 x + sin2 x = (sin x +

sin2 x)2=1

21. (b) 2 sin2θ+3(1–sin2θ)=3–sin2θ∴ leastvalue=3–1=2

22. (d) Let sec A+tanA=t

∴ 1sec tanA At

− =

By adding both 12sec A tt

= +

1 122

x tx t

+ = + ∴ t=2x or 12x


1. If5tanθ=4,then5sin 3cossin 2cos

θ− θθ+ θ

=

(a) 5/9 (b) 14/5(c) 9/5 (d) 5/14

2. (sec2 θ–1)(cosec2 θ–1)= [Karnataka CET-1998] (a) 0 (b) 1(c) secθ.cosecθ (d) sin2θ–cos2θ

3. The incorrect statement is [MNR-1993] (a) sinθ=1/5 (b) cosθ=1(c) sinθ=2 (d) tanθ=20

4. Which of the following relations is possible?(a) sinθ=5/3(b) tanθ=1002

(c) 2

2

1cos ( 1)1+

θ = ≠ ±−

p pp

(d) secθ=1/2

5. If sin θ + cos θ =m and sec θ + cosec θ=n, then n (m+1)(m–1)= [MPPET-1986](a) m (b) n(c) 2m (d) 2n

6. If cosecA+cotA= 112

, then tan A= [Roorkee-1995]

(a) 21/22 (b) 15/16(c) 44/117 (d) 117/43

7. If acosθ+bsinθ=m and asinθ–bcosθ=n, then a2+b2=(a) m+n (b) m2–n2 (c) m2+n2 (d) None of these

8. The value of 2(sin6 θ + cos6 θ) – 3(sin4 θ +cos4 θ)+1is [MPPET-1997; UPSEAT-2002](a) 2 (b) 0(c) 4 (d) 6

9. (sec A + tanA – 1) (secA – tanA + 1) –2 tan A= [Roorkee-1972](a) sec A (b) 2 sec A(c) 0 (d) 1

10. Iftanθ+sinθ=mandtanθ–sinθ=n, then [IIT-1970] (a) m2–n2=4mn(b) m2+n2=4mn(c) m2–n2=m2+n2

(d) 2 2 4− =m n mn

11. If0<x<πandcosx+sinx= 12

, then the

value of tan x is [MPPET-2009]

(a) 2 73−

(b) 4 7

3+

−

(c) 1 73+

−

(d) 2 73+

−


Important Instructions1. The answer sheet is immediately below the

worksheet.2. The worksheet is of 15 minutes.3. The worksheet consists of 15 questions. The

maximum marks are 45.4. UseBlue/BlackBallpointpenonlyforwriting

particulars/markingresponses.Useofpencilisstrictly prohibited.

1. Which of the following is equal to 1?(a) cos2θ–sin2θ(b) sec2θ–cosec2θ(c) cot2θ–tan2θ(d) sec2θ–tan2θ

2. The value of cos sinsec cosec

A AA A+ is

(a) sec2 A+tan2 A(b) sec2 A–tan2 A(c) cot2 A–cosec2 A(d) cosec2 A+cot2 A

3. 2

1 cossin+ θ

θ(a) 0 (b) 1

(c) 11 cos− θ

(d) 11 cos+ θ

4. 1 sin1 sin− θ+ θ

is equal to

(a) 0 (b) 1(c) secθ.tanθ (d) secθ–tanθ

5. The equation (a+b)2=4ab sin2θispossibleonly when(a) 2a=b (b) a =b(c) a=2b (d) None of these

6. sin6θ+cos6θ+3sin2θcos2θisequalto [MPPET-1995, 2002; DCE-2005](a) 0 (b) –1(c) 1 (d) None

7. If f(x)=cos2 x+sec2 x, then [MNR-1996]

(a) f (x) < 1 (b) f (x)=1(c) 1 < f (x) < 2 (d) f (x) ≥ 2

8. If sin θ + cos θ = a, then the value of |sinθ⋅cosθ|is [Pb. CET-92](a) 22 a− (b) 22 a+(c) 2 2a − (d) None

9. Which of the following is possible?

(a) cosθ= 75 (b) sinθ=

2 2

2 2

a ba b+−

(c) 5secθ=4 (d) tanθ=45

10. If cosx+cos2x=1,thenthevalueofsin12x+3sin10x+3sin8x+sin6x–1isequalto [VIT-2007](a) 2 (b) 1(c) –1 (d) 0

11. Ifsinθ+cosθ=1,thensinθcosθisequalto [Karnataka CET-1998](a) 0 (b) 1(c) 2 (d) 1/2

12. If(secα+tanα)(secβ+tanβ)(secγ+tanγ)= tan α tan β tan γ, then (sec α – tan α)(sec β–tanβ)(secγ–tanγ)isequalto [Krukshetra CEE-1998](a) cotαcotβcotγ(b) tanαtanβtanγ(c) cotα+cotβ+cotγ(d) tanα+tanβ+tanγ

13. The value of 6(sin6θ+cos6θ)–9 (sin4θ+cos4θ)+4is [MPPET-2001](a) –3 (b) 0(c) 1 (d) 3

14. If y=cos2 x+sec2 x, then [MP PET-2004](a) y ≤ 2 (b) y ≤ 1(c) y ≥ 2 (d) 1< y < 2

15. Ifsinθ+cosecθ=2, thevalueofsin10θ+cosec10θis

[MP PET-2004](a) 2 (b) 210

(c) 29 (d) 10


1. (d) from 2nd trigonometric identities 1+tan2θ=sec2θ⇒ sec2θ–tan2θ=1

2. (b) cos sinsec cosec

A AA A+

cos sin1 1

cos sin

A A

A A

= +

=cos2 A+sin2 A=1=sec2 A–tan2 A

3. (c) Given, 2

1 cossin+ θ

θ

2

1 cos1 cos+ θ

=− θ

[∴ sin2θ=1–cos2θ]

1 cos 1(1 cos )(1 cos ) 1 cos

+ θ= =

− θ + θ − θ

4. (d) Given

1 sin 1 sin1 sin 1 sin− θ − θ

×+ θ − θ

2 2

2 2

(1 sin ) (1 sin )1 sin cos− θ − θ

= =− θ θ

21 sincos− θ = θ

1 sin sec tancos cos

θ= − = θ− θ

θ θ

5. (b) Try yourself.

6. (c) sin6θ+cos6θ+3sin2θcos2θ=(sin2θ+cos2θ)3–3sin2θcos2θ+3sin2θcos2θ=1Short MethodPut θ = 0°,we get the values of expressionequal to 1.Again put θ = 45°, the value re-mains 1, it means that the expression is inde-pendentofθandisequalto1.

7. (d) f(x)=(secx–cosx)2+2⇒ f(x)≥2forallx.

8. (a)|sinθ–cosθ|2=(sinθ–csoθ)2=1–sin2θandsinθ+cosθ=a⇒ 1+sin2θ=a2

⇒ sin2θ=a2–1⇒ 1–sin2θ=2–a2

∴ |sinθ–cosθ| 22 a= −

9. (d) For option (a): cosθ cannot be greaterthan 1.For option (b): sinθ also cannot be greaterthan 1

For option (c): 4sec5

θ = is also not possible

assecθcannotbelessthen1.Foroption(d):tanθcanassumeany real value.∴ option (d) is correct.

10. (d) ∴ cos x+cos2 x=1⇒ cos x=1–cos2 x=sin2 x∴ sin12 x+3sin10 x+3sin8 x+sin6 x–1

1. a b c d 2. a b c d 3. a b c d 4. a b c d 5. a b c d




=cos6 x+3cos5 x+3cos4 x+cos3 x–1=(cos2 x+cosx)3–1=1–1=0

11. (a)Given,sinθ+cosθ=1By squaring both sides, we get(sinθ+cosθ)2=1⇒ sin2θ+cos2θ+2sinθcosθ=1

⇒ 1+2sinθcosθ=1

[∴ sin2θ+cos2θ=1]

⇒ 2sinθcosθ=0⇒sinθcosθ=0

12. (a) Given(secα+tanα)(secβ+tanβ)(secγ+tanγ)=tanαtanβtanγ ........(1)Let x = (sec α – tan α) (sec β – tan β) (secγ–tanγ) ................(2)By multiplying both equations (1) and (2), we get(sec2α–tan2α)(sec2β–tan2β)(sec2γ–tan2γ)=x.(tanαtanβtanγ)

1tan tan tan

x= =α β γ

∴ x=cotαcotβcotγ

13. (c) 6[(sin2 θ)3 + (cos2 θ)3] – 9 sin4 θ – 9 cos4θ+4=6[(sin2θ+cos2θ)(sin4θ–sin2θcos2θ)+(cos4θ)]–9sin4θ+4=–3(sin4θ+cos4θ+2sin2θcos2θ)+4=–3(sin2θ+cos2θ)2+4=–3+4=1

14. (c) y=cos2 x+sec2 x may be written asy=(cos2 x+sec2 x–2)+2or y=(cosx–secx)2+2As (cos x–secx)2is0or+ve∴ y=2+(positiveorzero)∴ y≥2.

15. (a) We have,sinθ+cosecθ=2

⇒ 1sin 2sin

θ+ =θ

1cosec sin

∴ θ = θ

⇒ sin2θ+1=2sinθ⇒ sin2θ–2sinθ+1=0⇒ (sinθ–1)2=0⇒sinθ=1Required value of sin10θ+cosec10θ

1010

1(1) 2(1)

= + =

trignometry basic questions

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