Q. 1. If and find the possible values of (α + β). (1978)
Ans.
Sol. We know tan
2. (a) Draw the graph of y =(sinx + cosx) from x = to
(b) If cos and α, β lies between 0 and , find tan2α. (1979)
Ans.
Sol. (a) Given: (sinx + cosx) sin ...(1)
Now, to draw the graph of we first draw
the graph of y = sin x and then on shifting it by we will obtain the required graph as shown in figure given below.
(b)
∴ tan 2α = tan [(α + β) + (α  β) ]
3. Given α + β  γ = π, prove that (1980)
Ans.
Sol. Given α + β – γ = π and to prove that
sin^{2} α + sin^{2}β – sin^{2}γ = 2sinα sinβ cosγ
L.H.S. = sin^{2}α + sin^{2}β – sin^{2}γ
[Using sin^{2}α – sin^{2}β = sin (A + B) sin (A – B)]
= sin^{2}α + sin(β + γ) sin (β – γ)
= sin^{2}α + sin(β + γ) sin (p – α) (∵ α + β – γ = π)
= sin^{2}α + sin(β + γ) sinα
= sinα (sinα + sin(β + γ))
= sinα [sin [π – (β  γ)] + sin(β + γ)]
= sinα [sin(β  γ) + sin(β + γ)]
= sinα [2 sinβ cosγ] = 2 sinα sinβ cosγ = R.H.S.
4. Given and f (x) = cos x – x (1 + x); find f (A). (1980)
Ans.
Sol.
f (x) = cos x – x (1+ x)
f '(x) = – sin x – 1– 2x < 0, ∀ x∈A
∴ f is a decreasing function.
5. For all θ in [0, π/ 2] show that, cos (sinθ) ≥ sin (cos q) . (1981  4 Marks)
Ans. Sol. We have
cosθ + sinθ
∴ cosθ + sinθ <π/ 2 ⇒ cosθ < π/ 2  sinθ ...(1)
As q ∈ [0,π / 2] in which sinθ increases.
∴ Taking sin on both sides of eq. (1), we get
sin (cosθ ) < sin ( π /2 –sinθ )
sin (cosθ ) < cos (sinθ )
⇒ cos (sinθ ) > sin (cosθ ) ....(1)
Hence the result.
6. Without using tables, prove that
(sin 12°) (sin 48°) (sin 54°) = (1982  2 Marks)
Ans. Sol. L.H.S. = sin 12° sin 48° sin 54° = [2 sin 12° cos 42°] sin 54°
Now we know that sin 54°
∴ We get, =
7. Show that 16cos (1983  2 Marks)
Ans. Sol. We know that,
cos A cos 2A cos 4A . . . . cos 2^{n}
(where A = 2π/15)
8. Find all the solution of 4 cos ^{2} x sin x  2 sin^{2} x= 3 sinx (1983  2 Marks)
Ans. Sol. Given eq. is,
4 cos^{2} x sin x – 2 sin^{2} x = 3 sin x
⇒ 4 cos^{2} x sin x – 2 sin^{2 }x – 3 sin x = 0
⇒ 4 (1 – sin^{2} x) sin x – 2 sin^{2} x – 3 sin x = 0
⇒ sin x [ 4 sin^{2} x + 2 sin x – 1] = 0
⇒ either sin x = 0 or 4 sin^{2} x + 2 sin x – 1 = 0
If sin x = 0 ⇒ x = np
⇒ If 4 sin^{2} x + 2 sin x – 1 = 0 ⇒ sin x =
If sin x =
then
If sin x == sin(54^{0})
then x
Hence,
where n is some integer
9. Find the values of x∈(–π, +π) which satisfy the equation
(1984  2 Marks)
Ans. Sol. The given equation is
The values of
10. Prove that tanα + 2 tan 2α + 4 tan 4α + 8 cot 8α = cotα (1988  2 Marks)
Ans. Sol. We know that tan
⇒ cot α – tan α = 2 cot 2α
Now we have to prove
tanα + 2 tan 2α + 4 tan 4α + 8 cot 8α = cot α
LHS
tan α + 2 tan 2α + 4 tan 4α + 4 (2 cot 2 . 4α)
= tan α + 2 tan 2α + 4 tan 4α + 4 (cot 4α  tan 4α) [Using (1)]
= tan α + 2 tan 2α + 4 tan 4α + 4 cot 4α  4 tan 4α
= tan α + 2 tan 2α + 2 (2 cot 2. 2 a )
= tan α + 2 tan 2α + 2 (cot α – tan 2α)
= tan α + 2 tan 2α + 2 (2 cot 2αtan 2α ) [Using (1)]
= tan α + 2 cot 2α
= tan α + (cot αtan α) [Using (1)]
= cot α = RHS.
11. ABC is a triangle such that sin(2A + B) = sin (C – A) = – sin (B + 2C) =
If A, B and C are in arithmetic progression, determine the values of A, B and C. (1990  5 Marks)
Ans. Sol. Given that in ΔABC, A, B and C are in A.P.
∴ A + C = 2B
also A + B + C = 180° ⇒ B + 2B = 180° ⇒ B = 60°
Also given that, sin (2A + B) = sin (C – A) = – sin (B + 2C) =
⇒ sin (2A + 60°) = sin (C – A) = – sin (60 + 2C) = ..(1)
From eq. (1), we have
sin (2A + 60°) = ⇒ 2A + 60° = 30° , 150°
but A can not be –ve
∴ 2A + 60° = 150° ⇒ 2A = 90 ° ⇒ A = 45°
Again from (1) sin (60° + 2C) = 
⇒ 60° + 2C = 210° or 330°
⇒ C = 75° or 135°
Also from (1) sin (C – A) =⇒ C – A = 30°, 150°
For A = 45°, C =75° or 195° (not possible) ∴ C = 75°
Hence we have A = 45° , B = 60°, C = 75°
12. If exp {(sin^{2}x + sin^{4}x + sin^{6}x + ............... ∞ ) In 2} satisfies the equation x^{2}– 9x + 8 = 0, find the value of
(1991  4 Marks)
Ans. Sol. Let y = exp [sin^{2} x +sin^{4 }x +sin^{6 }x + . . . . ∞ ] ln 2
As y satisfies the eq.
x^{2} – 9x + 8 = 0 ∴ y^{2} – 9y + 8 = 0
⇒ (y – 1) (y – 8) = 0 ⇒ y = 1, 8
⇒ tan 2 x = 0 or tan 2x = 3
⇒ tan x = 0 or
⇒ x = 0 or x = π/3, 2 π/3
But given that 0 < x < π/2 ⇒ x = π/3
Hence
13. Show that the value of wherever defined never lies between and 3. (1992  4 Marks)
Ans.
Sol.
Let
⇒ 3y – 3 tan^{2} x = 1– 3 tan^{2 }x
⇒ (y – 3) tan^{2} x = 3y – 1 ⇒
(L.H.S. being a prefect square)
⇒⇒ (3y  1) (y  3)>0
Thus y never lies between and 3
14. Determine the smallest positive value of x (in degrees) for which tan(x + 100°) = tan (x + 50°) tan(x) tan (x – 50°). (1993  5 Marks)
Ans. Sol. Given that, tan (x + 100°) = tan (x + 50°) tan x tan (x – 50°)
⇒ tan (x + 50°) tan (x – 50°)
⇒
⇒
Applying componendo and dividendo, we get
⇒
⇒ 2 sin (2x + 100°) cos 2x = – 2 sin 100° cos 100°
⇒ sin (4x + 100°) + sin 100° = – sin 200°
⇒ sin (4x + 10° + 90°) + sin (90° + 10°) = – sin (180 + 20°)
⇒ cos (4x + 10°) + cos 10° = sin 20°
⇒ cos (4x + 10°) = sin 20° – cos 10°
⇒ cos (4x + 10°) = sin 20° – sin 80°
= – 2 cos 50° sin 30° = – 2 cos 50°. = –cos 50° = cos 130°
⇒ 4x + 10° = 130° ⇒ x = 30°
15. Find the smallest positive number p for which the equation cos(p sin x) = sin(p cos x) has a solution x ∈ [0,2π]. (1995  5 Marks)
Ans.
Sol. Given that cosθ = sin φ
where θ = p sin x, φ = p cos x
Above is possible when both or
∴ p sin x = or p sin x =
and p cos x = or p cos x =
Squaring and adding,
only for least positive value or
16. Find all values of θ in the interval satisfying the equation (1996  2 Marks)
Ans.
Sol. Given :
or
Let us put tan^{2} θ = t
∴ (1– t) ( 1 + t) + 2^{t }= 0 or 1– t^{2} + 2^{t} = 0
It is clearly satisfied by t = 3.
as – 8 + 8 = 0 ∴ tan^{2} θ = 3
∴ p = ± π/3 in the given interval.
17. Prove that the values of the function do not lie between and 3 for any real x. (1997  5 Marks)
Ans.
Sol. Let
We have
(the expression is not defined if tan x = 0)
⇒ 3y – (tan^{2} x) y = 1– 3 tan^{2} x ⇒ 3y – 1= (y – 3) tan^{2} x
⇒
Since tan^{2} x > 0, we get (3y – 1) (y – 3) > 0
⇒ or y > 3
This shows that y cannot lie between and 3.
18. Prove that where n ≥ 3 is an integer. (1997  5 Marks)
Ans.
Sol. Expanding the sigma on putting k = 1, 2, 3, ......, n
S = (n – 1)
+ 1.cos (n – 1) ... (1)
We know that cosθ = cos (2π – θ)
Replacing each angle θ by 2π – θ in (1), we get
S = (n – 1) cos (n – 1) + (n – 2) cos (n – 2) + ......+ 1.cos by (1) .....(2)
Add terms in (1) and (2) having the same angle and take n common
∴ 2S=
Angles are in A.P. of
= n .1 cosπ = – n ∵ sin (π – θ) = sinθ ∴ S = – n/2
19. In any triangle ABC, prove that (2000  3 Marks)
Ans. Sol. We have, A + B + C = π
⇒
or
⇒
⇒
20. Find the range of values of t for which
(2005  2 Marks)
Ans. Sol. Given that, 2 sin t t ∈ [π/ 2,π/ 2]
This can be written as
(6 sin t – 5) x^{2} + 2 (1– 2 sin t) x – (1+ 2 sin t) = 0
For given equation to hold, x should be some real number, therefore above equation should have real roots i.e., D ≥ 0
⇒ 4 (1– 2 sin t)^{2} + 4 (6 sin t – 5) (1+ 2 sin t) ≥ 0
⇒
or
or
(Note that sin x is an increasing function from –π/2 to π/2)
∴ range of t is
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