trignometry basic questions

LECTURE

2

Trigonometric Functions

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

cos θ

tan θ

cot θ

sec θ

cos θ

sin θ

tan θ

1 − cos 2 θ

1 − sin 2 θ sin θ 2

1 − sin θ 1 − sin 2 θ sin θ 1 2

1 − sin θ

tan θ

1 + cot θ

1

cot θ 2

1 + tan θ

cos θ 2

1 − cos θ 1 cosθ

p b p ,cos θ = , tan θ = , h h b

h h b cosec θ = ,sec θ = ,cot θ = ; p b p

1 + cot θ 1 cot θ

1 tan θ

cotθ

2

1 + tan θ

1 cosec θ

1 secθ

2

tanθ

cosec θ

sec 2 θ − 1 sec θ

2

1 + tan θ

cosθ 1 − cos 2 θ cos θ

sec θ 1

2

2. T-RATIOS (OR TRIGONOMETRIC FUNCTIONS) sin θ =

cot θ

cosec 2θ − 1 cosec θ 1

sec 2 θ − 1

cosec 2θ − 1

1

cosec 2 θ − 1

sec 2 θ − 1

1 + cot 2 θ cot θ

cosec θ

secθ

cosec 2 − 1

B

p h O

p +b =h 2

2

2

θ b

A

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

A.18 Trigonometric Functions

3. FOLLOWING ARE SOME OF THE FUNDAMENTAL TRIGONOMETRIC IDENTITIES (i) sin θ =

1 cosec θ

1 or cosec θ = sin θ (ii) cos θ =

1 sec θ

or sec θ =

1 cos θ

1 (iii) cot θ = tan θ or tan θ = (iv) tan θ =

1 cot θ

sin θ cos θ

cos θ sin θ 2 (v) sin θ + cos2 θ = 1 or sin2 θ = 1 – cos2 θ or cos2 θ = 1 – sin2 θ (vi) sec2 θ – tan2 θ = 1 or cot θ =

(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 + cosec x | ≥ 2 (xi) |cos x + sec x | ≥ 2 (xii) |tan x + cot x | ≥ 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) (ii) (iii) (iv) (v)

sin2 θ + cos4 θ = sin4 θ + cos2 θ = 1 – sin2 θ cos2 θ sin4 θ + cos4 θ =1 – 2 sin2 θ cos2 θ sin6 θ + cos6 θ = 1 – 3 sin2 θ cos2 θ sin4 θ – cos4 θ = 1 – 2 cos2 θ sin8 θ – cos8 θ = (sin2 θ – cos2 θ) (1 – 2 sin2 θ cos2 θ) (vi) sec2 θ + cosec2 θ = sec2 θ – cosec2 θ = tan2 θ + cot2 θ + 2 (vii)

1 1 + sin θ = sec θ + tan θ = sec θ − tan θ 1 − sin θ

(viii)

1 1 − sin θ = sec θ − tan θ = sec θ + tan θ 1 + sin θ

or 1 + tan2 θ – sin2 θ

1 sec θ + tan θ (vii) cosec2 θ – cot2 θ = 1 or 1 + cot2 θ = cosec2 θ 1 or cosec θ − cot θ = cosec θ + cot θ or sec θ − tan θ =

4. MAXIMUM AND MINIMUM VALUES OF TRIGONOMETRICAL FUNCTIONS (i) – 1≤ sin x ≤ 1

(ix)

1 1 + cos θ = cosecθ + cot θ = cosecθ − cot θ 1 − cos θ

(x)

1 1 − cos θ = cosecθ − cot θ = cosecθ + cot θ 1 + cos θ

(xi)

1 − sin θ cos θ = = sec θ − tan θ cos θ 1 + sin θ

(xii)

1 − cos θ sin θ = = cosec θ − cot θ sin θ 1 + cos θ

(xiii)

1 − sin θ cos θ = = sec θ − tan θ cos θ 1 + sin θ

(xiv)

1 − cos θ sin θ = = cosec θ − cot θ sin θ 1 + cos θ

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

Trigonometric Functions A.19

1. Show that

(1 + tan θ + cot θ)(sin θ − cos θ) sec3 θ − cosec3θ

3. If

sin A cos A = m and = n , fi nd the value of sin B cos B

tan B; n2 < 1 < m2.

Solution

 sin θ cos θ  1+ +  (sin θ − cos θ) (1 + tan θ + cot θ)(sin θ − cos θ)  cosSolution θ sin θ  = L.H.S. = 1 1 A sec3 θ − cosec3θ − sin = m ⇒ sin A = m sin B .........(1) Given 3 3 cos θ sinsin θB  sin θ cos θ  1+ + (sin θ − cos θ) cos A (1 + tan θ + cot θ)(sin θ − cos θ)  cos θ sin θ  = n ⇒ cos A = n cos B ..........(2) and = 3 3 cos B 1 1 sec θ − cosec θ − cos3 θ sin 3 θ By squaring (1) and (2) and adding, we get =

=

(sin θ cos θ + sin 2 θ + cos 2 θ)(sin θ − cos θ)  sin 3 θ − cos3 θ  sin θ cos θ   3 3  sin θ cos θ  (sin 3 θ − cos3 θ) sin 3 θ cos3 θ sin θ cos θ (sin 3 θ − cos3 θ)

( (a 2 + b 2 + ab)(a − b) = a 3 − b3 ) = sin 2 θ cos 2 θ = R.H.S.

2. If cos2 α – sin2 α = tan2 β, then show that tan2 α = cos2 β – sin2 β. Solution

Given, cos2 α – sin2 α = tan2 β ⇒

cos 2 α − sin 2 α sin 2 β = cos 2 α + sin 2 α cos 2 β

cos 2 α + sin 2 α cos 2 β ⇒ = cos 2 α − sin 2 α sin 2 β (∴ cos 2 α + sin 2 α = 1)

Applying componendo and dividendo, we get cos 2 α + sin 2 α + cos 2 α − sin 2 α cos 2 β + sin 2 β = cos 2 α + sin 2 α − cos 2 α + sin 2 α cos 2 β − sin 2 β

2cos 2 α 1 ⇒ = 2sin 2 α cos 2 β − sin 2 β

1 = m 2 sin 2 B + n 2 cos 2 B

⇒

1 sin 2 B = m2 + n 2 (Dividing by cos2 2 cos B cos 2 B B)

⇒ sec 2 B = m 2 tan 2 B + n 2 ⇒ 1 + tan 2 B = m 2 tan 2 B + n 2 ⇒ 1 − n 2 = (m 2 − 1) tan 2 B ⇒ tan 2 B = ⇒ tan B = ±

1 − n2 m2 − 1

1 − n2 m2 − 1

4. Eliminate θ between cosecθ − sin θ = a, sec θ − cos θ = b. Solution

Given cosecθ − sin θ = a

.............(1)

and sec θ − cos θ = b

.............(2)

From (1), From (2),

1 cos 2 θ − sin θ = a ⇒ =a sin θ sin θ

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

1 sin 2 θ − cos θ = b ⇒ =b cos θ cos θ

..............(4) Squaring (3) and multiplying by (4), we get cos3 θ = a 2b ⇒ cos θ = ( a 2b)1/3

............(5)

sin 2 α ⇒ = cos 2 β − sin 2 β cos 2 α

Squaring (4) and multiplying by (3), we get

⇒ tan α = cos β − sin β, as desired.

sin 3 θ = ab 2 ⇒ sin θ = (ab 2 )1/3

2

2

2

............(6)

A.20  Trigonometric Functions

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

Solution

Given x = r cos θ cos φ, y = r cos θ sin φ and Squaring and adding, we get 2

sin 8 A cos8 A 1 + = a3 b3 ( a + b)3 Solution

z = r sin θ. 2

sin 4 A cos 4 A 1 + = , then prove that a b a+b

7. If

2

(a + b) (b sin4 A + a cos 4A) – ab = 0

x +y +z = r 2 cos 2 θ cos 2 φ + r 2 cos 2 θ sin 2 φ + r 2 sin 2 θ

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

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

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

= r 2 (sin 2 θ + cos 2 θ) = r 2 [∴ sin 2 θ + cos 2 θ = 1]  sin α cos α = m and = n , than prove that 6. If sin β cos β tan α = ±

m 1 − n2 2 ; n < 1 < m2 . n m2 − 1

Given

sin α =m sin β

............(1)

and

cos α =n cos β

.............(2)

sin α From (1), sin β =  m

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

cos α  n

.............(4)

sin 2 α cos 2 α + m2 n2

⇒ m 2 n 2 = n 2 sin 2 α + m 2 cos 2 α

Divide throughtout by cos2 α ⇒

m 2 n 2 n 2 sin 2 α m 2 cos 2 α = + cos 2 α cos 2 α cos 2 α 2

2

2

⇒ m n (1 + tan α) = n tan α + m

=

( a + b) 1 = 4 ( a + b) ( a + b)3

8. If sin x + sin2 x + sin3 x = 1, then 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 – 4 cos2 x + cos4 x) = cos4 x ⇒ cos6 x – 4 cos4 x + 8 cos2 x = 4.

Solution 2

⇒ m 2 n 2 − m 2 = n 2 tan 2 α − m 2 n 2 tan 2 α m2 n2 − 1 ⇒ tan 2 α = 2 × n 1 − m2 m 1 − n2 . n m2 − 1



sinθ.

⇒ m 2 n 2 + m 2 n 2 tan 2 α = n 2 tan 2 α + m 2

⇒ tan 2 α = ±

sin 8 A cos8 A 1 a4 1 b4 + = 3. + 3. 3 3 4 a b a ( a + b) b ( a + b) 4

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

⇒ m 2 n 2 sec 2 α = n 2 tan 2 α + m 2 2 2

cos 2 A sin 2 A 1 = = b a a+b

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

Squaring (3) and (4) and adding, we get 1=

or ∴

Solution

and from (2), cos β =

or (a cos2 A – b sin 2 A)2 = 0

(Proved)

8 sinθ = 4 + cosθ ⇒ 8 sinθ – 4 = cosθ ⇒ (8 sinθ – 4)2 = cos2θ ⇒ ⇒ ⇒ ⇒

(8 sinθ – 4)2 = 1 – sin2θ 64 sin2θ + 16 – 64 sinθ = 1 – sin2θ 65 sin2θ – 64 sinθ + 15 = 0 65 sin2 θ – 39 sinθ – 25 sinθ + 15 = 0

Trigonometric Functions A.21

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

⇒ sin θ =

5 3 or . 13 5

sin θ cos θ cos θ sin θ = + sin θ − cos θ cos θ − sin θ sin θ cos θ

⇒ sin θ =

3 in fi rst quadrant 5

=

and sin θ =

5 in second quadrant 13

sin 2 θ cos 2 θ + cos θ (sin θ − cos θ) sin θ(cos θ − sin θ)

=

sin 3 θ − cos3 θ cos θ sin θ (sin θ − cos θ)

⇒ (13 sinθ – 5) (5 sin θ – 3) = 0

10. Prove that, tan θ cot θ + = sec θ cosecθ + 1 1 − cot θ 1 − tan θ Solution

L.H.S. =

tan θ cot θ + 1 − cot θ 1 − tan θ

EXERCISE 1 1. Prove that, sec4 θ – sec2 θ = tan4 θ + tan2 θ sin θ 1 + cos θ + = 2cosec θ 2. Prove that, 1 + cos θ sin θ 3. Prove that,

cosec θ cosec θ + = 2sec 2 θ cosec − 1 cosec + 1

4. Prove that, (cosec θ – sin θ) (sec θ – cos θ) (tan θ + cot θ) = 1 3 5. If tan θ = , then fi nd all the trigonometric 4 ratios. 6. If sec θ + tan θ = 4, fi nd the values of sin θ, cos θ, sec θ and tan θ. 7. Prove that,

1 + cos x = (cosec x + cot x) 2 1 − cos x

8. If sin x + sin2 x = 1, then prove that cos 2 x + cos4 x = 1

[a 3 − b3 = (a − b)(a 2 + b 2 + ab)] =

(sin θ − cos θ) (sin 2 θ + cos 2 θ + sin θ cos θ) cos θ sin θ (sin θ − cos θ)

=

1 + sin θ cos θ = sec θ cosec θ + 1 sin θ cos θ

9. If secθ + tanθ = p, obtain the values of secθ, tanθ in terms of p. 10. Prove that, sin θ cot θ + sin θ cosec θ = 1 + cos θ 11. Prove that, tan 2 θ cosec 2θ 1 + = 2 2 tan θ − 1 sec θ − cosec 2θ sin 2 θ − cos 2 θ 12. Prove that, (1 − tan x) 2 + (1 − cot x) 2 = (sec x − cosecx) 2 . EXERCISE 2 1. Prove that tan2 θ – sin2 θ = tan2 θ – sin2 θ 2. Prove that sec θ − tan θ = 1 − 2sec θ tan θ + 2 tan 2 θ sec θ + tan θ 3. Prove that

1 + cos θ = (cosec θ + cot θ) 1 − cos θ

A.22 Trigonometric Functions

4. Prove that

1 − cos θ sin θ = sin θ 1 + cos θ

5. Prove that cot4 θ + cot2 θ = cosec4 θ – cosec2 θ 6. Prove that (Sin A + cos A) (tan A + cot A) = sec A + cosec A 7. Prove that cos Acosec A − sin A sec A = cosec A − sec A cos A + sin A 8. Prove that 2 sin A + cos A = 1 + sin A 2

4

4

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

1 + cos x = (cosec x + cot x) 2 1 − cos x

11. If sin x + sin2 x = 1, prove that cos2 x + cos4 x = 1 12. If sec θ + tan θ = p, obtain the values of sec θ, tan θ in terms of p.

EXERCISE 1

EXERCISE 2 15 8 15 17 sin θ = ,cos θ = , tan θ = ,sec θ = 17 17 8 8 3 4 4 5 5 p2 + 1 p2 −1 p2 −1 5. sin θ = ,cos θ = ,cot θ = ,cosecθ = ,sec θ = , tan θ = and sinθ = 2 12. = sec θ = 5 15 5 3 17 4 8 3 15 2p 2p p +1 sin θ = ,cos θ = , tan θ = ,sec θ = 17 17 8 8 2 2 2 3 4 4 55 5 and psin−θ p − 1 p +1 1 sin θ = ,cos θ = ,cot θ = ,cosec cosecθθ== ,sec , θ= sec θ2= , tan θ = 2 and sinθ = 2 5 5 3 33 4 2 p +1 p +1 p − 12 p p 2−p1 , tan θ = and sinθ= 2 9. sec θ = 4 4 5 5 2p 2p p +1 ,cot θ = ,cosecθ = ,sec θ = 5 3 3 4 p2 + 1 p2 −1 p2 −1 sec θ = , tan θ = and sin θ = 15 2 p8 , tan θ = 152,sec p θ = 17 p2 +1 6. sin θ = ,cos θ = 17 17 8 8

2sin a 1 − cos α + sin α then 1 + cos α + sin α 1 + sin α is equal to [UPSEAT, 99] (a) 1/y (b) y (c) 1 – y (d) 1 + y

1. If y =

=

(1 + sin α) 2 − cos 2 α (1 + sin α)(1 + cos α + sin α)

=

(1 + sin 2 α + 2sin α) − (1 − sin 2 α) (1 + sin α)(1 + cos α + sin α)

=

2sin α(1 + sin α) (1 + sin α)(1 + cos α + sin α)

=

2sin α =y 1 + cos α + sin α

Solution

(b)

1 − cos α + sin α 1 + sin α =

1 − cos α + sin α 1 + cos α + sin α . 1 + sin α 1 + cos α + sin α

Trigonometric Functions  A.23

2. If a cos3α + 3a cosα sin2α = m and a sin3α + 3a cos2α sinα = n, then (m + n)2/3 + (m – n) 2/3 is equal to (a) 2a2 (c) 2a2/3

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

(b) 2a1/3 (d) 2a3

(a) x =

y +1 y −1

(b) x =

y −1 y +1

(c) y =

1+ x 1− x

Solution

(c) From the given relations, we get m + n = a cos3 α + 3a cos α sin2 α + 3a cos2 α sin α + a sin3 α = a (cosα + sin α)3 similarly, m – n = a(cos α – sin α)3 ∴ (m + n)2/3 + (m – n)2/3 = a2/3  [(cos α +sin α)2 + (cos α – sin α)2]

(d) xy + x – y + 1 = 0 Solution

(b, c, d) We have x =

By multiplying, we get

= a2/3 [2 (cos2 α + sin2 α)] = 2a2/3



3. (m + 2) sinθ + (2m – 1) cos θ = 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) tan θ + (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) tan2 θ + (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

1 − sin φ 1 + cos φ ,y = cos φ sin φ

xy =

(1 − sin φ)(1 + cos φ) cos φ sin φ

⇒ xy + 1 1 − sin φ + cos φ − sin φ cos φ + sin φ cos φ = cos φ sin φ =



1 − sin φ + cos φ and cos φ sin φ

x− y =





(1 − sin φ)sin φ − cos φ(1 + cos φ) cos φ sin φ

=

sin φ − sin 2 φ − cos φ − cos 2 φ cos φ sin φ

=

sin φ − cos φ − 1 = −( xy + 1) cos φ sin φ

Thus, xy = x − y + 1 = 0 ⇒ x =

y −1 and y +1



1+ x 1− x

y=

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

(b) 2, 1/4 (d) 4, 1/8

Solution

⇒ (3 tanθ – 4) [(1 – m2) tanθ + 2m] = 0

(a) 1 + sin 2 A = 3sin A cos A

Which is true if tanθ =4/3 or tanθ = 2m/ (m2 – 1)

⇒ 2sin 2 A − 3sin A cos A + cos 2 A = 0

⇒ (sin 2 A + cos 2 A) + sin 2 A = 3sin A cos A ⇒ 2 tan 2 A − 3tan A + 1 = 0

A.24  Trigonometric Functions

⇒ 2sin 2 A − 3sin A cos A + cos 2 A = 0 ⇒ 2 tan 2 A − 3tan A + 1 = 0

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

⇒ tan A – 1 = 0, 2 tan A – 1 = 0 ⇒ tan A = 1, 1/2 6. For what values of x is the equation 2sinθ = x + 1/x is valid? [NDA-2006]

(a) (b) (c) (d)

9. If sin x + sin2 x = 1, then cos6 x + cos 12 x + 3 cos10x + 3 cos8 x is equal to  [Pb. CET-2002; MPPET-2006] (a) 1 (b) cos3 x sin3 x (c) 0 (d) ∞ Solution

(a) ⇒ ⇒ ∴

x = ±1 all real values of x –1 < x < 1 x > 1 and x < –1

Solution

(a) x2 – 2xsinθ + 1 = 0 ⇒ x=



10. The value of the expression

[Bihar EE-1990] sin 2 y 1 + cos y sin y 1− + − is equal to 1 + cos y sin y 1 − cos y



2sin θ ± 4sin 2 θ − 4 2

= sin θ ± sin 2 θ − 1

(a) 0 (c) sin y

Which is valid when sinθ = ±1 Then x = ±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 (c) θ, ϕ

(d) The given value = 1 – (1 – cos y) + cos y + 0 = cos y

Solution

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

(b) 2 (d) cosθ

Solution

(b) 1 (d) cos y

Solution

(b) r, ϕ (d) r, θ

(c) x2 + y 2 + z2 = r2 sin2θ cos2ϕ + r2 sin2θ sin2ϕ + r2 cos2θ = r2 sin2θ (cos2ϕ + sin2ϕ) + r2 cos2θ = r2 (sin2θ + cos2θ) = r2.

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

11.

1 − cos 2 y − sin 2 y = sin y (1 − cos y )

2sin θ tan θ(1 − tan θ) + 2sin θ sec 2 θ is equal (1 + tan θ) 2

[Roorkee-1975]

to

(a)

sin θ 1 + tan θ

(b)

2sin θ 1 + tan θ

(c)

2sin θ (1 + tan θ) 2

(d) None of these

Solution

(b) Given that expression

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

=

2sin θ {tan θ (1 − tan θ) + sec 2 θ} (1 + tan θ) 2



=

(1 − cos 2 θ) (1 − sin 2 θ) (1 + tan 2 θ) × × cos θ sin θ tan θ

=

2sin θ {tan θ − tan 2 θ + 1 + tan 2 θ} (1 + tan θ) 2



=

sin 2 θ.cos 2 θ.sec 2 θ.cos θ = 1. cos θ.sin θ.sin θ

=

2sin θ 1 + tan θ

Trigonometric Functions  A.25

12. If cotθ + tanθ = m and secθ – 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

Solution

15. If cos x = tan y, cos y = tan z, cos z = tan x; than prove that sin x = sin y = sin z = 2 sin 18o.

As given,

1 + tan θ = m ⇒ 1 + tan 2 θ = m tan θ tan θ

⇒ sec θ = m tanθ ................. (i) 2

and secθ – cosθ = n ⇒ sec θ – 1 = n secθ 2

(d) Given that, y = sin2 θ + cosec2 θ ∴ y = (sin θ – cosec θ)2 + 2 ⇒ y ≥ 2, θ ≠ 0

Solution

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

cos 2 z tan 2 x = 1 − cos 2 z 1 − tan 2 x

or 1 + cos 2 x =

sin 2 x cos 2 x − sin 2 x

⇒ tan2θ = n secθ ⇒ tan4θ = n2 sec2 θ = n2 m tanθ {by (i)} ⇒ tan3θ = n2m, ⇒ tanθ = (n2m)1/3

[∵tanθ ≠ 0) ................ (ii)

Also, sec2θ = m tanθ = m(n2m)1/3

or

c hanging to sin x, we get (2 – sin2 x) (1 – 2 sin2 x) = sin2x

or 2 sin4 x – 6 sin2 x + 2 = 0 or sin4 x – 3 sin2 x+1=0

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

∴ sin 2 x =

⇒ m(mn2)1/3 – n (nm2)1/3 = 1 4 xy is true, if and only if ( x + y)2  [AIEEE-2002] (a) x – y ≠ 0 (b) x = –y (c) x = y (d) x ≠ 0, y ≠ 0

13. sin 2 θ =

Solution (c) Since, θ ≤ 1

⇒ x2 + y2 + 2xy – 4xy ≥ 0

it is > 1. or sin 2 x =

6 − 2 5  5 −1  =   4  2 

3+ 5 as 2

2

5 −1 ( 5 − 1) = 2. = 2sin18° 2 4 By symmetry, we can say that sin x = sin y = sin z = 2 sin18o.

16. Show that, cos (sin θ) > sin (cos θ), 0 ≤ θ ≤ π .

⇒ (x – y) ≥ 0 2

Which is true for all real values of x and will be meaningless.

We have rejected the value

∴ sin x =

  4 xy 4 xy ≤ 1  sin 2 θ = given  ⇒⇒ 2 2 ( x + y)   ( x + y)

y provided x + y ≠ 0, otherwise,

3± 9− 4 3± 5 3− 5 = = 2 2 2

4 xy ( x + y)2

14. If y = sin2 θ + cosec2 θ, θ ≠ 0, then  [AIEEE-2002] (a) y = 0 (b) y ≤ 0 (c) y ≥ –2 (d) y ≥ 2

2 [IIT-JEE-1981]

 Solution sin θ + cos θ ≤ 2 < ⇒ sin θ <

π − cos θ 2

π 2

π  ⇒ cos (sin θ) > cos  − cos θ  = sin (cos θ) 2 

A.26 Trigonometric Functions

2t , then cos θ is equal to 1+ t2 2t 2t (a) (b) 1− t2 1+ t2

1. If sin θ =

(c) 2.

1− t2 1+ t2

(d)

1+ t2 1− t2

(b) 1 (d) cos θ + sin θ

1 3. If for real values of x, cos θ = x + , then x [MPPET-1996] (a) θ is an acute angle (b) θ is a right angle (c) θ is an obtuse angle (d) No values of θ is possible 4. The equation sec2 θ = possible when

4 xy ( x + y)2

is only

[MPPET-1986; IIT-1996] (a) x = y (c) x > y

20 21 (d) ± 21 29 3 3 8. If x = a cos θ, y = b sin θ, 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 (c) ±

sin θ cos θ + is equal to 1 − cot θ 1 − tan θ [Karnataka CET-1998] (a) 0 (c) cos θ – sin θ

20 , cos θ will be 21 [MPPET-1994] 20 1 (b) ± (a) ± 41 21

7. If tan θ =

(b) 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) sin 1 = sin 1o π (d) sin1 = sin1° 180 6. If sin θ + cosec θ = 2, then sin2 θ + cosec2 θ is equal to [MPPET-1992; MNR-1990; UPSEAT-2002] (a) 1 (b) 4 (c) 2 (d) None of these

1 is equal to x [MPPET-1986] (b) 2 sec θ (d) 2 tan θ

9. If x = sec θ + tan θ, then x + (a) 1 (c) 2

10. If (1 + sin A) (1 + sin B) (1 + sin C) = (1 – sin A) (1 – sin B) (1 – sin C), 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 sin θ1 + sin θ2 + sin θ3 = 3, then cos θ1 + cos θ2 + cos θ3 is equal to [EAMCET-1994] (a) 3 (b) 2 (c) 1 (d) 0 12. If tan θ – cot θ = a and sin θ + cos θ = b, then (b2 – 1) 2 (a2 + 4) is equal to [WB JEE-1979] (a) 2 (b) – 4 (c) ± 4 (d) 4 13. If tan θ = then

x sin φ y sin θ and tan φ = , 1 − x cos φ 1 − y cos θ

x is equal to y

(a)

sin φ sin θ

(c)

sin φ 1 − cos θ

[MPPET-1991] sin θ sin φ sin θ (d) 1 − cos φ (b)

Trigonometric Functions A.27

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

14. If sec θ + tan θ = p, then tan θ is equal to [MPPET-1994] 2p p2 −1 (a) 2 (b) p −1 2p (c)

p2 +1 2p

15. If p =

(d)

2p p2 +1

20. If sinx + sin2x = 1, then cos8x + 2 cos6 x + cos4x is equal to (a) 0 (b) –1 (c) 2 (d) 1

2sin θ cos θ and q = , then 1 + cos θ + sin θ 1 + sin θ

(a) pq = 1 (c) q – p = 1

[MPPET-2001] q (b) = 1 p (d) q + p = 1

16. The minimum value of 9 tan2 θ + 4 cot2 θ is (a) 13 (b) 9 (c) 6 (d) 12 17. The maximum value of 4 sin2 x + 3 cos2 x is [Karnataka CET-2003] (a) 3 (b) 4 (c) 5 (d) 7

W &

tan A is (a) 3x (c)

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

BC2 = (1 – t2) ⇒ BC = 1 – t2

∴

cos θ =

BC 1 − t 2 = AC 1 + t 2

Second Method θ 2 tan 2t 2 . If t = tan θ sin θ = = 2 θ 2 1+ t 1 + tan 2 2

x 2

1 , then the value of sec A + 4x [MPPET-2010] x (b) 3

(d) 2x

θ 2 2 = 1− t cos θ = θ 1+ t2 1 + tan 2 2 1 − tan 2

W %

(b) 2 (d) 5

22. If sec A = x +

∴ 

AC2 = AB2 + BC2 (1 + t2)2 = 4t2 + BC2

(a) 1 (c) 3



$

2t 1+ t2 By Pythagoras theorem

1. (c) Given sin θ =

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

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

2. (d) Given

=

=

⇒

sin θ cos θ + 1 − cot θ 1 − tan θ

sin θ cos θ + cos θ sin θ 1− 1− sin θ cos θ sin θ cos θ + sin θ − cos θ cos θ − sin θ sin θ cos θ sin 2 θ cos 2 θ + sin θ − cos θ cos θ − sin θ

A.28  Trigonometric Functions

⇒

sin 2 θ cos 2 θ − sin θ − cos θ sin θ − cos θ

⇒

sin 2 θ − cos 2 θ sin θ − cos θ

⇒

(sin θ + cos θ)(sin θ − cos θ) (sin θ − cos θ)

⇒ sin θ + cos θ 3. (d) Given equation is cos θ = x + 1/x or x2 – x cos θ + 1 = 0 for real value of x By B2 > 4AC ∵ A = 1, B = –cos θ, C = 1 cos2 θ > 4 (1) (1) or |cos θ| ≤ 2 Which is impossible because |cos θ| < 1 ∴ No real value of θ is possible. 4 xy ≥1 4. (a) sec θ ≥ 1 ∀ θ ⇒ ( x + y)2 2 ⇒ 4xy ≥ (x + y) 2

⇒ (x + y)2 – 4xy ≤ 0 ⇒ (x – y)2 ≤ 0. But (x – y)2 ≥ 0 ⇒ (x – y) = 0 ⇒ x = y 2

5. (a) 1 radian = mately)

180 degrees = 57°  (approxiπ

⇒ sinx is increasing function

1°  < 1c , sin (1°) < sin (1c)

6. (c) sin θ + cosec (θ) = 2

By sec2 θ = 1 + tan2 θ ∴

sec 2 θ = 1 +

or cos 2 θ =

400 441

441 21 or cos θ = ± 841 29

8. (c) Given x = a cos3 θ, y = b sin3 θ x y = cos3 θ, = sin 3 θ a b taking cube root on both sides. ⇒

1/3

⇒

x   a

1/3

 y = cos θ,   b

= sin θ

Now square and add 2

2

1 1       x  3  +   y  3  = sin2 θ + cos2 θ   a     b      

⇒

x   a

9. (b) x +

2/3

 y +  b

=1

1 1 = (sec θ + tan θ) + x sec θ + tan θ

= (sec θ + tan θ) + ⇒

2/3

x+

(sec θ − tan θ) sec 2 θ − tan 2 θ

1 = 2 sec(θ) x

10. (b) Multiplying both sides by (1 – sin A) (1 – sin2 B) (1 – sin2 C) We have  (1 – sin2 A) (1 – sin2 B) (1 – sin2 C)

= (1 – sin A)2 (1 – sin B)2 (1 – sin C)2

⇒ ( 1 – sin A) (1 – sin B) (1 – sin C) = ± cos A cos B cos C

1 ⇒ sin θ + =2 sin θ

Similarly, (1 + sin A) (1 + sin B) (1 + sin C)

⇒ sin2 θ – 2 sinθ + 1 = 0



⇒ (sin θ – 1)2 = 0 ⇒ sin θ = 1 ⇒ θ = π/2 ∴ sin2 (θ) + cosec2 (θ) = 1 + 1 = 2 20 7. (c) Given tan θ = Squaring both sides 21 400 tan 2 θ = 441

= ± cos A cos B cos C

11. (d) Given sin θ1 + sin θ2 + sin θ3 = 3 ⇒ sin θ1 = 1 = sin θ2 = sin θ3 ⇒ θ1 = θ2 = θ3 = π/2 ⇒ cos θ1 = cos θ2 = cos θ3 = 0 ⇒ cos θ1 + cos θ2 + cos θ3 = 0

12. (d) Given, tan θ – cot θ = a sin θ + cos θ = b

…….. (1) ……….. (2)

Trigonometric Functions  A.29

16. (d) Since, A.M. ≥ G.M.

(2) ⇒ b2 = 1 + 2 sin θ cos θ ⇒ b2 – 1 = sin (2θ)

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

2

2

= tan2 θ + cot2 θ – 2 = a2 + 4 = (tan θ + cot θ)2  sin θ cos θ  + = a2 + 4 =    cos θ sin θ 

2

 sin θ + cos θ   2  = =   sin θ cos θ   sin 2θ  2

=

2

17. (b) f(x) = 4 sin2 x + 3 cos2 x = sin2 x + 3 and 0 ≤ |sin x| ≤ 1 ∴ maximum value of sin2 x + 3 is 4.

2

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

4 ⇒ (a2 + 4) (b2 – 1)2 = 4 (b 2 − 1)

13. (b) tan θ =

⇒ cos2 x + sin2 x + 2 sin x cos x + k sin x cos x – 1 = 0∀x ⇒ (k + 2) sin x cos x = 0∀x

x sin φ 1 − x cos φ

⇒ k + 2 = 0 ⇒ k = –2

⇒ (1 – x cos ϕ) tan θ = x sin ϕ ⇒

x=

tan θ  sin φ + cos φ tan φ

Similarly, tan φ =

………. (1)

y sin θ 1 − y sin θ ………. (2)

(1) x tan θ  sin θ + cos θ tan φ  ⇒ = .  (2) y tan φ  sin φ + cos φ tan φ  sin θ cot φ + cos θ sin φ cot θ + cos φ sin θ(cot φ + cot θ) sin θ = = sin φ(cot θ + cot φ) sin φ

sec θ − tan θ =

(1) – (2) ⇒ p −

Now, a2 = tan2 θ + cot2 θ – 2 tan θ . cot θ = tan2 θ + cot2 θ – 2 ⇒

=

………….. (1)

1  p

 sin θ cos θ  a2 + 4 =  +   cos θ sin θ 

2

 sin 2 θ + cos 2 θ   2  = =   sin θ cos θ   sin 2θ 

=

⇒

19. (c) x sin3 α + y cos3 α = sin α . cos α …(1) x sin α – y cos α = 0 …….. (2)

⇒ a2 + 4 = (tan θ + cot θ)2

tan φ ⇒ y=  sin θ + cos θ tan θ

14. (b) sec θ + tan θ = p

Therefore, 9 tan 2 θ + 4cot 2 θ ≥ 9 tan 2 θ× 4 cot 2 θ 2 ⇒ 9 tan 2 θ + 4cot 2 θ ≥ 2 × 36 = 12

(1) ⇒ a = tan θ + cot θ – 2 tan θ cot θ 2

………… (2) 2

1 p −1 = 2 tan θ ⇒ tan θ = p 2p

2sin θ cos θ + 15. (d) p + q = 1 + sin θ + cos θ 1 + sin θ

2

4 ⇒ (a2 + 4) (b2 – 1)2 = 4 (b 2 − 1)

20. (d) We have, sin x + sin2 x = 1 ⇒ sin x = 1 – sin2 x ⇒ sin x = cos2 x ⇒ cos8 x + 2 cos6 x + cos4 x ⇒ sin4 x + 2 sin3 x + sin2 x = (sin x + sin2 x)2 = 1 21. (b) 2 sin2 θ + 3 (1 – sin2 θ) = 3 – sin2 θ ∴ least value = 3 – 1 = 2 22. (d) Let sec A + tan A = t

=

2sin θ + 2sin 2 θ + cos θ + sin θ cos θ + cos 2 θ (1 + sin θ + cos θ)(1 + sin θ)

=

(1 + sin θ + cos θ) + (sin 2 θ + sin θ + sin θ cos θ) (1 + sin θ + cos θ)(1 + sin θ)

By adding both 2sec A = t +

=

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

2x +

∴

sec A − tan A =

1 t

1 t

1 1 1 = t +   ∴  t = 2x or 2x t 2x

A.30 Trigonometric Functions

1. If 5 tan θ = 4, then (a) 5/9 (c) 9/5

5sin θ − 3cos θ = sin θ + 2cos θ (b) 14/5 (d) 5/14

2. (sec2 θ – 1) (cosec2 θ – 1) = [Karnataka CET-1998] (a) 0 (b) 1 (c) sec θ.cosec θ (d) sin2 θ – cos2 θ 3. The incorrect statement is [MNR-1993] (a) sin θ = 1/5 (b) cos θ = 1 (c) sin θ = 2 (d) tan θ = 20 4. Which of the following relations is possible? (a) sin θ = 5/3 (b) tan θ =1002 1 + p2 ( p ≠ ±1) (c) cos θ = 1 − p2 (d) sec θ =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 11 6. If cosecA + cot A = , then tan A = 2 [Roorkee-1995] (a) 21/22 (b) 15/16 (c) 44/117 (d) 117/43

7. If acos θ + bsin θ = m and asin θ – bcos θ = 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) + 1 is [MPPET-1997; UPSEAT-2002] (a) 2 (b) 0 (c) 4 (d) 6 9. (sec A + tan A – 1) (sec A – tan A + 1) – 2 tan A = [Roorkee-1972] (a) sec A (b) 2 sec A (c) 0 (d) 1 10. If tan θ + sin θ = m and tan θ – sin θ = n, then [IIT-1970] (a) m2 – n2 = 4mn (b) m2 + n2 = 4mn (c) m2 – n2 = m2 + n2 (d) m 2 − n 2 = 4 mn 1 , then the 2 [MPPET-2009]

11. If 0 < x < π and cos x + sin x = value of tan x is

(a)

2− 7 3

(c) −

1+ 7 3

(b) −

4+ 7 3

(d) −

2+ 7 3

Trigonometric Functions A.31

Important Instructions 1. 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. Use Blue/Black Ball point pen only for writing particulars/marking responses. Use of pencil is strictly prohibited. 1. Which of the following is equal to 1? (a) cos2 θ – sin2 θ (b) sec2 θ – cosec2 θ (c) cot2 θ – tan2 θ (d) sec2 θ – tan2 θ cos A sin A + is sec A cosec A sec2 A + tan 2 A sec2 A – tan2 A cot2 A – cosec2 A cosec2 A + cot2 A

2. The value of (a) (b) (c) (d) 3.

4.

1 + cos θ sin 2 θ (a) 0 1 (c) 1 − cos θ 1 − sin θ is equal to 1 + sin θ (a) 0 (c) sec θ . tan θ

1 1 + cos θ

(b) 1 (d) sec θ – tan θ

5. The equation (a + b)2 = 4ab sin2 θ is possible only when (a) 2a = b (b) a = b (c) a = 2b (d) None of these 6. sin6 θ + cos6 θ + 3 sin2 θ cos2 θ is equal to [MPPET-1995, 2002; DCE-2005] (a) 0 (b) –1 (c) 1 (d) None 7. If f(x) = cos2 x + sec2 x, then

(b) f (x) = 1 (d) f (x) ≥ 2

8. If sin θ + cos θ = a, then the value of |sin θ⋅cos θ| is [Pb. CET-92] (a)

2 − a2

(b)

2 + a2

(c)

a2 − 2

(d) None

9. Which of the following is possible? 7 a 2 + b2 (b) sin θ = 2 2 (a) cos θ = 5 a −b (c) 5 sec θ = 4 (d) tan θ = 45 10. If cosx + cos2x = 1, then the value of sin12x + 3 sin10x + 3 sin8x + sin6x – 1 is equal to [VIT-2007] (a) 2 (b) 1 (c) – 1 (d) 0 11. If sin θ + cos θ = 1, then sin θ cos θ is equal to [Karnataka CET-1998] (a) 0 (b) 1 (c) 2 (d) 1/2

(b) 1 (d)

(a) f (x) < 1 (c) 1 < f (x) < 2

[MNR-1996]

12. If (sec α + tan α) (sec β + tan β) (sec γ + tan γ) = tan α tan β tan γ, then (sec α – tan α) (sec β – tan β) (sec γ – tan γ) is equal to [Krukshetra CEE-1998] (a) cot α cot β cot γ (b) tan α tan β tan γ (c) cot α + cot β + cot γ (d) tan α + tan β + tan γ 13. The value of 6(sin6 θ + cos6 θ) – 9 (sin4 θ + cos4 θ) + 4 is [MPPET-2001] (a) –3 (b) 0 (c) 1 (d) 3 14. If y = cos2 x + sec2 x, then (a) y ≤ 2 (c) y ≥ 2

[MP PET-2004] (b) y ≤ 1 (d) 1< y < 2

15. If sin θ + cosec θ = 2, the value of sin10 θ + cosec10 θ is [MP PET-2004] (a) 2 (b) 210 (c) 29 (d) 10

A.32 Trigonometric Functions

a a a a a

1. 2. 3. 4. 5.

b b b b b

c c c c c

6. 7. 8. 9. 10.

d d d d d

a a a a a

1. (d) from 2nd trigonometric identities

1 + tan θ = sec θ

⇒

sec2 θ – tan2 θ = 1

2

= cos2 A + sin2 A = 1 = sec2 A – tan2 A

[∴ sin2 θ = 1 – cos2 θ]

4. (d) Given

b b b b b

c c c c c

d d d d d

Short Method Put θ = 0°, we get the values of expression equal to 1. Again put θ = 45°, the value remains 1, it means that the expression is independent of θ and is equal to 1. f(x) ≥ 2 for all x.

8. (a) |sin θ – cos θ|2 = (sin θ – cso θ)2 = 1 – sin 2θ and sin θ + cos θ = a ⇒ 1 + sin 2θ = a2 ⇒ sin 2θ = a2 – 1 ⇒ 1 – sin 2θ = 2 – a2 ∴ |sin θ – cos θ| = 2 − a 2

1 − sin θ 1 − sin θ × 1 + sin θ 1 − sin θ (1 − sin θ) 2 (1 − sin θ) 2 = 1 − sin 2 θ cos 2 θ

 1 − sin θ  =    cos θ 

a a a a a

6. (c) sin6 θ + cos6 θ + 3 sin2 θ cos2 θ

⇒

1 + cos θ 1 = = (1 − cos θ)(1 + cos θ) 1 − cos θ

=

11. 12. 13. 14. 15.

d d d d d

7. (d) f(x) = (sec x – cos x)2 + 2

1 + cos θ 3. (c) Given, sin 2 θ 1 + cos θ 1 − cos 2 θ

c c c c c

= (sin2 θ + cos2 θ)3 – 3 sin2 θ cos2 θ + 3 sin2 θ cos2 θ = 1

2

cos A sin A cos A sin A + = + 2. (b) 1 1 sec A cosec A cos A sin A

=

b b b b b

2

1 sin θ = − = sec θ − tan θ cos θ cos θ 5. (b) Try yourself.

9. (d) For option (a): cosθ cannot be greater than 1. For option (b): sinθ also cannot be greater than 1 4 For option (c): sec θ = is also not possible 5 as sec θ cannot be less then 1. For option (d): tanθ can assume any real value. ∴

option (d) is correct.

10. (d) ∴ cos x + cos2 x = 1 ⇒ ∴

cos x = 1 – cos2 x = sin2 x sin12 x + 3 sin10 x + 3 sin8 x + sin6 x – 1

Trigonometric Functions  A.33

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

= 6[(sin2 θ + cos2 θ) (sin4 θ – sin2 θ cos2 θ) + (cos4 θ)] – 9 sin4 θ + 4

11. (a) Given, sin θ + cos θ = 1 By squaring both sides, we get

= –3(sin4 θ + cos4 θ + 2 sin2 θ cos2 θ) + 4

(sin θ + cos θ)2 = 1 ⇒ sin2 θ + cos2 θ + 2 sin θ cos θ = 1 ⇒ 1 + 2 sin θ cos θ = 1 

[∴ sin2 θ + cos2 θ = 1]

⇒ 2 sin θ cos θ = 0 ⇒ sin θ cos θ = 0 (sec α + tan α) (sec β + tan β) (sec γ + tan γ) ........ (1)

Let x = (sec α – tan α) (sec β – tan β) (sec γ – tan γ) ................ (2) By multiplying both equations (1) and (2), we get (sec2 α – tan2 α) (sec2 β – tan2 β) (sec2 γ – tan2 γ) = x.(tan α tan β tan γ) =x=

1 tan α tan β tan γ

∴ x = cot α cot β cot γ

= –3(sin2 θ + cos2 θ)2 + 4 = –3 + 4 = 1 14. (c) y = cos2 x + sec2 x may be written as y = (cos2 x + sec2 x – 2) + 2 or y = (cos x – sec x)2 + 2 As (cos x – sec x)2 is 0 or +ve ∴ y = 2 + (positive or zero)

12. (a) Given = tan α tan β tan γ

13. (c) 6[(sin2 θ)3 + (cos2 θ)3] – 9 sin4 θ – 9 cos4 θ + 4

∴ y ≥ 2. 15. (a) We have, sin θ + cosec θ = 2 1 =2 ⇒ sin θ + sin θ

1   ∴ cosec θ = sin θ 

⇒ sin2 θ + 1 = 2 sin θ ⇒ sin2 θ – 2 sin θ + 1 = 0 ⇒ (sin θ – 1)2 = 0 ⇒ sin θ = 1 Required value of sin10 θ + cosec10 θ = (1)10 +

1 =2 (1)10