Eamcet Mathematics - Trigonometric Functions , Identities and Equation Chapter Sample Question Paper With Answer Key
Exam Duration: 60 Mins Total Questions : 50
The value of \({ sin6 }^{ \circ }.{ sin }42^{ \circ }.{ sin66 }^{ \circ }.{ sin78 }^{ \circ }\)is
- (a)
\(\frac { 1 }{ 13 } \)
- (b)
\(\frac { 1 }{ 14 } \)
- (c)
\(\frac { 1 }{ 15 } \)
- (d)
\(\frac { 1 }{ 16 } \)
\({ sin6 }^{ \circ }.{ sin }42^{ \circ }.{ sin66 }^{ \circ }.{ sin78 }^{ \circ }\)
\(={ sin6 }^{ \circ }.{ cos }48^{ \circ }.{ cos24 }^{ \circ }.{ cos12 }^{ \circ }\)
\({ =sin6 }^{ \circ }\frac { { 2 }^{ 3 }.{ si12 }^{ \circ }.{ cos }12^{ \circ }.{ cos24 }^{ \circ }.{ cos48 }^{ \circ } }{ { 2 }^{ 3 }{ sin12 }^{ \circ } } \)
\(={ sin6 }^{ \circ }.\frac { { sin }96^{ \circ } }{ { 2 }^{ 3 }{ sin12 }^{ \circ } } \quad \quad \ \left[ \because \ sin2A=2sinA\quad cosA \right] \)
\(=\frac { 2{ sin }6^{ \circ }cos{ 6 }^{ \circ } }{ { 2 }^{ 4 }{ sin12 }^{ \circ } } =\frac { { sin }12^{ \circ } }{ { 2 }^{ 4 }{ sin12 }^{ \circ } } =\frac { 1 }{ 16 } \)
If \(cos\alpha +cos\beta =0\quad and\quad sin\alpha +sin\beta =0,\quad then\quad cos2\alpha +cos2\beta \) is equal to
- (a)
\(2cos\left( \alpha +\beta \right) \)
- (b)
\(-2cos\left( \alpha +\beta \right) \)
- (c)
\(3cos\left( \alpha +\beta \right) \)
- (d)
\(None\quad of\quad the\quad above\)
Given, cosα+cosβ=0 and sinα+sinβ=0
On squaring and subtracting both the equations, we get
( cosα+cosβ)2-(sinα+sinβ)2=0
⇒ (cos2α+sin2α +(cos2β-sin2β)+2[cosαcosβ-sinαsinβ]=0
⇒ cos2α+cos2β+2cos(α+β)=0
⇒ cos2α+cos2β=-2cos(α+β)
If \(cos(\theta +\phi )=mcos(\theta -\phi ),\) then \(\frac { 1-m }{ 1+m } cot\phi \) is equal to
- (a)
\(tan\theta \)
- (b)
\(-tan\theta \)
- (c)
\(2tan\theta \)
- (d)
\(None\quad of\quad these\)
\(\frac{1}{m}=\frac{cos(\theta-\phi)}{cos(\theta+\phi)}\)
On applying componendo and dividendo rule, we get
\(\frac{1-m}{1+m}=\frac{cos (\theta-\phi)-cos(\theta+\phi)}{cos(\theta-\phi)+cos(\theta+\phi)}\)
=\(\frac{2sin(\theta)sin(\phi)}{2cos(\theta)cos(-\phi)}\)=tanθtan\(\phi\)
∴ \((\frac{1-m}{1+m})cot\phi = tan\theta\)
The number of distinct solutions of \(secx+tanx=\sqrt { 3 } ,\) where \(0\le x\le 3\pi \) are
- (a)
1
- (b)
2
- (c)
3
- (d)
4
Given , secx+tanx=\(\sqrt{3}\)
⇒ 1+sinx=\(\sqrt{3}\) cosx
⇒ \(\sqrt{3}\)cosx-sinx=1
On dividing both sides by \(\sqrt{a^{2}+b^{2}}\) ,i.e., \(\sqrt{4}\)=2, we get
\(\frac{\sqrt{3}}{2} cosx-\frac{1}{2}sinx=\frac{1}{2}\)
\(\Rightarrow cos\frac{\pi}{6}cosx-sin\frac{\pi}{6}sinx=\frac{1}{2}\)
\(\Rightarrow cos(x+\frac{\pi}{6})=cos(\frac{\pi}{6})\Rightarrow x+\frac{\pi}{6}=2n\pi \pm \frac{\pi}{3}\)
\(\Rightarrow x=2n\pi+\frac{\pi}{6}, 2n\pi-\frac{\pi}{2}\)
When, x=2nπ-\(\frac{\pi}{2}\) , does not satisfies the equations
∴ Total number of solutions are 2.
If \(2{ sin }^{ 2 }\theta =3cos\theta ,\quad where\quad 0\le \theta \le 2\pi ,\) then find the value of \(\theta \) .
- (a)
\(\frac { 2\pi }{ 3 } ,\frac { 5\pi }{ 3 } \)
- (b)
\(\frac { \pi }{ 3 } ,\frac { 5\pi }{ 3 } \)
- (c)
\(\frac { 5 }{ 3 } ,\frac { 4\pi }{ 3 } \)
- (d)
\(None\quad of\quad these\)
Given, 2sin2θ=3 cosθ
∴ 2(1-cos2θ)=3cosθ
⇒ 2 cos2θ+3cosθ-2=0
⇒ (cosθ+2)(2cosθ-1)=0
⇒ cosθ=-2, cosθ=\(\frac{1}{2}\)
But cosθ ∊ [-1,1]
∴ cosθ ≠- -2
Now, cosθ=\(\frac{1}{2}\) and
as θ=∈[0,2π]
∴ θ=\(\frac{\pi}{3}, \frac{5\pi}{3}\)
If \(0
- (a)
\(\frac { (4-\sqrt { 7 } ) }{ 3 } \)
- (b)
\(-\frac { (4+\sqrt { 7 } ) }{ 3 } \)
- (c)
\(\frac { (1+\sqrt { 7 } ) }{ 4 } \)
- (d)
\(\frac { (1-\sqrt { 7 } ) }{ 4 } \)
Gives, cosx+sinx=\(\frac{1}{2}\)
∴ \(\frac{1-tan^{2}\frac{x}{2}}{1+tan^{2}\frac{x}{2}}+\frac{2tan\frac{x}{2}}{1+tan^{2}\frac{x}{2}}=\frac{1}{2}\)
Put tan\(\frac{x}{2}=t\)
∴ \(\frac{1+t^{2}}{1+t^{2}}+\frac{2t}{1+t^{2}}=\frac{1}{2}\)
⇒3t2-4t-1=0 ⇒ \(t = {2 \pm \sqrt{7} \over 3}\)
As 0<x<ㅠ ⇒ 0<\(\frac{x}{2}<\frac{\pi}{2}\)
So, tan\(\frac{x}{2}\) is positive.
ஃ t=tan\(\frac{x}{2}\)= \( {2 + \sqrt{7} \over 3}\)
Now, tanx=\(\frac{2tan\frac{x}{2}}{1-tan^{2}\frac{x}{2}}=\frac{2t}{1-t^{2}}\Rightarrow tanx=\frac{2(\frac{2+\sqrt{7}}{3})}{1-(\frac{2+\sqrt{7}}{3})^{2}}\)
\( \Rightarrow tanx=-3(\frac{2+\sqrt{7}}{1+2\sqrt{7}})\Rightarrow tanx=-(\frac{4+\sqrt{7}}{3})\)
| tan x + see x| = |tan x| + Isee x|, x ∈ [0, 2π], if and only if x belongs to the interval
- (a)
[0, π]
- (b)
\([0,{\pi\over 2})\cup({\pi\over 2}, \pi]\)
- (c)
\([0,{3\pi\over 2})\cup({3\pi\over 2}, 2\pi]\)
- (d)
(π, 2π]
If |f(x) + g(x)| =|f(x)| + |g(x)|
⇔ f(x), g(x) > 0
tan x see x>'0
\(⇒{sin\ x\over cos^2x}\ge0\)
⇒ sin x>0
But cos x ≠ 0
\(x\ ∈[0,\pi]\left\{\pi\over 2\right\}\)
or \([0,{\pi\over 2})\cup({\pi\over 2}, \pi]\)
If 2 tan2 x - 5 see x is equal to 1 for exactly 7 distinct values of x ∈ \(\left[0,{n\pi\over 2}\right], n\epsilon\ N\) then the greatest value of n
- (a)
6
- (b)
12
- (c)
13
- (d)
15
2 tan2 x - 5 see x = 1
⇒2 (sec2 x-1) - 5 see x = 1
⇒2 sec2 x - 5 see x - 3 = 0
⇒(see x - 3) (2 see x + 1) = 0
|sec x| > 1
sec x = 3
Which gives two values of x in [0,2π], (2π, 4π], (4π, 6π] and one value in [ 6π,61π+ 3π/2]
\(6\pi+{3\pi\over 2}={15\pi\over 2}\)
Hence, greatest value of n is 15.
The general solution of the trigonometrical equation sin x + cos x = 1 for n = 0, ± 1, ± 2 ... is given by
- (a)
x = 2nπ, n∈I
- (b)
x = 2nπ+\(π\over 2\), n∈I
- (c)
x = nπ+(-1)n\(π\over 4\)-\(π\over 4\), n∈I
- (d)
none of these
Since, sin x + cos x = 1
\(⇒ {1\over \sqrt2}sin\ x+{1\over \sqrt2}cos\ x={1\over \sqrt2}\)
\(\Rightarrow\ sin\left(x+{\pi\over 4}\right)=sin{\pi\over 4}\)
\(\Rightarrow\ x+{\pi\over 4}=n\pi+(-1)^n{\pi\over 4}\)
\(x=n\pi+(-1)^n{\pi\over4}-{\pi\over 4},n\epsilon I\)
The number of solutions of the equation 2 (sin4 2x + cos4 2x) + 3 sin2 x cos2 x = a is
- (a)
0
- (b)
1
- (c)
2
- (d)
3
Since, 2 (sin4 2x + cos4 2x) + 3 sin2 x cos2 x = 0
⇒ 2{(sin2 2x + cos2 2X)2 - 2sin2 2x cos2 2x}+\({3\over 4}sin^22x=0\)
⇒ 2 {1 - 2 sin2 2 x (1 - sin2 2 x)} +\({3\over 4}sin^22x=0\)
⇒ 2 (1 - 2 sin 2 2 x + 2 sin 4 2 x) + \({3\over 4}sin^22x=0\)
⇒ 8-16sin22x+16sin42x+3sin22x=0
⇒ 16sin42x-13sin22x+8=0
Here, B2 - 4AC < 0
Hence, no solution.
ie, no. of solutions = 0
The complete solution of the equation 7 cos2 x + sin x cos x - 3 = a is given by
- (a)
\(n\pi+{\pi\over 2}(n\epsilon I)\)
- (b)
\(n\pi-{\pi\over 2}(n\epsilon I)\)
- (c)
\(n\pi+tan^{-1}\left(4\over 3\right)(n\epsilon I)\)
- (d)
\(n\pi+{3\pi\over 4},k\pi+tan^{-1}\left(4\over 3\right)(k,n\epsilon I)\)
Since, 7 cos2 x + sin x cas x - 3 = 0
Dividing by cos2 x, then
7 + tan x - 3 sec2x = 0
⇒ 7 + tan x - 3 (1 + tan2 x) = 0
⇒ 3 tan2 x - tan x - 4 = 0
⇒ (tan x + 1)(3 tan x - 4) = 0
tan x = - 1 and tan x =\({4\over 3}\)
Hence, \(x=n\pi+{3\pi\over 4}\ and\ x=k\pi+tan^{-1}\left(4\over 3\right)\)
where (k, n ∈ I)
The most general values of x for which sin x + cos x =\(\underset{a\epsilon\ R}{min}\) {1, a2 - 4a + 6} are given by
- (a)
2nπ, n ∈ N
- (b)
2nπ+\(\pi\over 2\) , n ∈ N
- (c)
nπ+(-1)n\({\pi\over 4}-{\pi\over 4}\) n ∈ N
- (d)
none of these
sin x + cos x =\(\underset{a\epsilon\ R}{min}\) {1, a2 - 4a + 6}
a2 - 4a + 6 = (a - 2)2 + 2 > 2
for all a
sin x + cas x = 1
\(sin\left(\pi+{\pi\over 4}\right)={1\over \sqrt2}\)
\(x+{\pi\over 4}=n\pi+(-1)^n{\pi\over 4}\)
x=nπ+(-1)n\({\pi\over 4}-{\pi\over 4}\)
If x ∈ (0, 1) the greatest root of the equation sin 2π x =√2 cos π X is
- (a)
1/4
- (b)
1/2
- (c)
3/4
- (d)
none of these
Given x ∈ (0,1)
and sin 2πx =√2 cos πx
⇒cos πx(2sin πx-√2)=0
⇒ \(cos\pi x\left(sin\pi x-{1\over \sqrt2}=0\right)\)
cos πx = 0 and sin π x =\(1\over \sqrt2\)
\(\pi x=n\pi+{\pi\over 2}\ and\ \pi x=n\pi+(-1)^n{\pi\over 4}\)
or \(x=n+{1\over 2}\ and\ x=n+
(-1)^n{1\over 4}n\epsilon I\)
for n = 0 and for n = 0, 1
\(x={1\over 2},x={1\over 4},{3\over 4}\)
Hence, greatest root =\({3\over 4}\)
If max {5 sin θ + 3 sin (θ - α)} = 7, then the set of possible values of a is (θ ∈ R)
- (a)
\(\{x:x=2n\pi\pm{\pi\over 3}, n\epsilon I\}\)
- (b)
\(\{x:x=2n\pi\pm{2\pi\over 3}, n\epsilon I\}\)
- (c)
\(\left[{\pi\over 3},{2\pi\over 3}\right]\)
- (d)
none of these
Since, 5 sin θ + 3 sin (θ - α) = 5 sin θ + 3 (sin θ cas α - cas θ sin α)
(5 + 3 cos α) sin θ - 3 sin α cos θ
max {5 sin θ + 3sin (θ - α)}
=\(\sqrt{\{ (5+3cos\alpha)^2 +9sin^2\alpha\}}\)
\(=\sqrt{(34+30cos\alpha)}\)
\(\sqrt{(34+30cos\alpha)}=7\)
\(⇒\ cos\alpha={49-34\over 30}\)
\(⇒ cos\alpha={1\over 2}=cos{\pi\over 3}\)
\(⇒\alpha =2n\pi\pm{\pi\over 3}n\epsilon I\)
The solution of the inequality log1/2 sin x > log1/2 cos x in [0, 2π] is
- (a)
\(x\epsilon\left(0,{\pi\over 2}\right)\)
- (b)
\(x\epsilon\left(0,{\pi\over 8}\right)\)
- (c)
\(x\epsilon\left(0,{\pi\over 4}\right)\)
- (d)
none of these
If sin x>0
X ∈ (0, 1t)
cos x > 0
\(x\epsilon\left(0,{\pi\over 2}\right)\cup\left({3\pi\over 2},2\pi\right)\)
From Eqs. (i) and (ii)
\(x\epsilon\left(0, {\pi\over 2}\right)\)
Now, log1/2 sin x > Iog1/2 cas x
\(sin\ x<cos\ x\ in \left(0,{\pi\over 2}\right)\)
\(x\epsilon\left(0,{\pi\over 4}\right)\)
The solution set of the inequality \(cos^2\theta<{1\over 2}\) is
- (a)
\(\left\{\theta:(8n+1){\pi\over 4}<\theta<(8n+3){\pi\over 4},n\epsilon I\right\}\)
- (b)
\(\left\{\theta:(8n-31){\pi\over 4}<\theta<(8n-1){\pi\over 4},n\epsilon I\right\}\)
- (c)
\(\left\{\theta:(4n+1){\pi\over 4}<\theta<(4n+3){\pi\over 4},n\epsilon I\right\}\)
- (d)
none of these
\(cos^2\theta<{1\over 2}\)
\(2cos^2\theta-1<0\)
cos 2θ<0
-cos2θ>0
cos(π+2θ)> cos\(\pi\over 2\)
\(2n\pi-{\pi\over 2}<\pi+2\theta>2n\pi+{\pi\over 2},n\epsilon I\)|
\(2n\pi-{3\pi\over 2}<2\theta<2n\pi-{\pi\over 2}\)
or\( n\pi-{3\pi\over 4}<\theta\ 2n\pi-{\pi\over 2}\)
Replacing n by n + 1, then
\((4n+1){\pi\over 4}<\theta<(4n+3){\pi\over 4}, n\epsilon I\)
2 sin x cos 2x = sin x, if
- (a)
\(x=n\pi+{\pi\over 6}(n\epsilon I)\)
- (b)
\(x=n\pi-{\pi\over 6}(n\epsilon I)\)
- (c)
\(x=n\pi(n\epsilon I)\)
- (d)
\(x=n\pi+{\pi\over 2}(n\epsilon I)\)
sin x (2cos2x-l) = 0
sin x = 0, cas 2x =\(1\over 2\)
x = nπ, 2x = 2nπ\(\pm{\pi\over 3}\)
or x = nπ, x = 2nπ\(\pm{\pi\over 6}n\epsilon I\)
The equation \(2sin\left(x\over 2\right)cos^2x-2sin^2x=cos^2x-sin^2\) has a root for which
- (a)
sin 2x = 1
- (b)
sin 2x = - 1
- (c)
\(cos\ x={1\over 2}\)
- (d)
\(cos\ 2x=-{1\over 2}\)
\(2sin{x\over 2}(cos^2x-2sin^2x)-(cos^2x-sin^2)=0\)
\(\left(2sin{x\over 2}-1\right)(cos^2x-sin^2x)=0\)
\(\left(2sin{x\over 2}-1\right)cos2x=0\)
\(sin{x\over 2}={1\over 2}\)an d cas 2x = 0
x= 60° and x = ± 450
sin x + cos x = 1+ sin x cos x, if
- (a)
\(sin\left(x+{\pi\over 4}\right)={1\over \sqrt2}\)
- (b)
\(sin\left(x-{\pi\over 4}\right)={1\over \sqrt2}\)
- (c)
\(cos\left(x+{\pi\over 4}\right)={1\over \sqrt2}\)
- (d)
\(cos\left(x-{\pi\over 4}\right)={1\over \sqrt2}\)
Since, sin x + cos x = 1 + sin x cos x
sin x (1 - cas x) - (1- cas x) = 0
⇒ (sin x-1) (1- cas x) = 0
⇒ sin x = 1 and cas x = 1
Then, \(x={\pi\over v2}\)or x = 0
An equation of the form f(sin x ± cas x, ± sin x cas x) = 0
can be solved by changing variable. I
Let sin x ± cas x = t
sin2 x + cos2 ± 2 sin x cas x = t2
\(±sin\ x\ cos\ x=\left({t^2-1\over 2}\right)\)
Hence, reduce the given equation into
\(f=\left(t{t^2-1\over 2}\right)=0\)
If sin2 x + cos4 X = sin x cosx, then x is
- (a)
nπ, n∈ I
- (b)
\((6n+1){\pi\over 6}n\epsilon I\)
- (c)
\((4n+1){\pi\over 4},n\epsilon I\)
- (d)
none of these
sin4 x + cos4 X = sin x cas x
⇒ (sin2 x + cos2 X)2 - 2 sin2 x cos2 X = sin x cas x
⇒ 1 - 2 sin2 x cos2 X = sin x cas x
Let sin x cas x =⋋, then
1-2⋋2=⋋
2 ⋋2+ ⋋-I = 0
(⋋ + 1)(2. ⋋ - 1) = 0
\(⋋=-1{1\over 2}\)
⇒ \(sin\ x\ cos\ x=-1,{1\over 2}\)
sin 2x = - 2, 1
sin 2x ≠ -2
sin 2x = 1
or \(2x = 2n\pi+{\pi\over 2}, n\epsilon I\)
\(x=(4n+1){\pi\over 4}, n\epsilon I\)
When ever the terms on the two sides of the equation are of different nature, then equations are known as Non standard form, some of them are in the form of an ordinary equation but can not be solved by standard procedures.
Non standard problems require high degree of logic, they also require the use of graphs, inverse properties of functions, in equalities .
If \(0\le x\le2\pi\ and\ 2^{cosec^2x}\sqrt{\left({1\over 2}y^2-y+1\right)}\le\sqrt2\) then number of ordered pairs of (x, y) is
- (a)
1
- (b)
2
- (c)
3
- (d)
infinitely many
\(2^{cosec^2x}\sqrt{\left({1\over 2}y^2-y+1\right)}\le\sqrt2\)
⇒ \(2^{cosec^2x}\sqrt{(y^2-2y+2)}\le\sqrt2\)
⇒\(2^{cosec^2x}.\sqrt{\{(y-1)^2+1\}}\le\sqrt2\)
Since, cosec2x > 1 for all real x
\(2^{cosec^2}\ge2\)
Also (y -1)2 + 1 ≥ 1
\(⇒ \sqrt{(y-1)^2+1}\ge2\)
From Eqs. (ii) and (iii),
\(2^{cosec^2x}.\sqrt{(y-1)^2+1}\ge2\)
Now, Eqs. (i) and (iv), equality holds only when
\(2^{cosec^2x}=2\ and\sqrt{\{(y-1)^2+1\}}=1\)
or cosec+e = 1 and (y - 1)2 + 1 = 1
⇒ sin x = ± 1 and y = 1
⇒ x = π / 2, 3π / 2 and y = 1
Hence, the solution of the given inequality is
\((x,y)\equiv\left({\pi\over 2},1\right),\left({3\pi\over 2},1\right)\)
The angles of a triangle are in A.P. and the number of degrees in the least to the number of radians in the greatest is 60: \(\pi\), find the angles in degrees.
- (a)
30o, 60o, 90o
- (b)
40o, 60o, 90o
- (c)
30o, 30o, 120o
- (d)
20o, 130o, 30o
Let the angles of the triangle be (a - d)o, ao, (a + d)o, where d > 0 .....(i)
then (a - d) + a + (a + d) = 180 \(\Rightarrow\) a = 60
\(\therefore\) From (i), the angles are (60 - d)o, 60o, (60 + d)o
Now, the least angle = (60 - d)o and the greates angle = (60 + d)o
= (60 + d) x \(\frac{\pi}{180}\) radian
By the given condition, we have
\(\frac{60-d}{\frac{\pi}{180}(60+d)}=\frac{60}{\pi}\Rightarrow\frac{180(60-d)}{(60+d)}=60\)
\(\Rightarrow\) 180 - 3d = 60 + d \(\Rightarrow\) 4d = 120 \(\Rightarrow\) d = 30
\(\therefore\) From (i), the angles are (60 - 30)o, 60o, (60 + 30)o
i.e., 30o, 60o, 90o.
Find the radius of the circle in which a central angle of 60° intercepts an arc of length 37.4 cm (use \(\pi=\frac{22}{7}\))
- (a)
37.5 cm
- (b)
32.8 cm
- (c)
35.7 cm
- (d)
34.5 cm
Here l = 37.4 cm and \(\theta=60^o=\frac{60\pi}{180}\) radian = \(\frac{\pi}{3}\)
Hence, by r = \(\frac{1}{\theta}\), we have
r = \(\frac{37.4\times 3}{\pi}=\frac{37.4\times 3\times 7}{22}\) = 35.7 cm
Find the length of an arc of a circle of radius 3cm, if the angle subtended at the centre is 300 .
- (a)
1.50cm
- (b)
1.35cm
- (c)
1.57cm
- (d)
1.20cm
Let l be the length of the arc. We know that Angle \(\theta ={1\over r}\)where θ is in radian
Given r=3cm θ=300=30 x \({\pi\over 180}={\pi\over 6}rad\)
On putting the values of r and θ, we get
\({\pi\over 6}={l\over 3}⇒l={\pi\over 2}={3.14\over 2}=1.57cm\)
A circular wire of radius 3cm is cut and bent so as to lie along the circumference of a hoop whose radius is 48cm. Find the angle in degrees which is subtended at the centre of hoop
- (a)
21.50
- (b)
23.50
- (c)
22.50
- (d)
24.50
Length of wire = 2πx3=6πcm and r=48cm is the radius of the hoop. Therefore the angle θ (in radian) subtended at the centre of the hoop is given by
\(\theta={Arc\over Radius }={6\pi\over 48}={\pi\over8}=22.5^0\)
Find the circular measure of the following angle 750
- (a)
\(\pi\over 3\)
- (b)
\(5\pi\over 12\)
- (c)
\(3\pi\over 4\)
- (d)
\(2\pi\over 7\)
Now 1800=π radian
\(⇒\ 1^0={\pi\over 180}radian⇒75^0={75\pi\over 180}={5\pi\over 12}\ radian\)
Find the circular measure of the following angle -22030'
- (a)
-\(\pi\over 8\)
- (b)
\(\pi\over 12\)
- (c)
-\(\pi\over 12\)
- (d)
\(\pi\over 4\)
since 1800=π radian, therefore, \(1^0={\pi\over 180}\)radian
\(⇒(-22^030')=\left(-22{1\over 2}\right)^0={\pi\over 180}\left(-{45\over 2}\right)radian\)
\(=-{\pi\over 8} radian\left(∵30'=\left({30\over 60}\right)^0=\left(1\over 2\right)\right)\)
Find the radian measure of 5200
- (a)
13π/9
- (b)
26π/9
- (c)
17π/9
- (d)
6π/9
Required radian measure = \({\pi\over 180}\times \) Degree Measure
\(={\pi\over 180}\times520={26\pi\over 9}\)
If tan A +cot A =4, then tan4A+cot4A is equal to
- (a)
110
- (b)
191
- (c)
80
- (d)
194
tan A+cot A=4 ...(i)
Squaring (i) both sides, we get
tan2A+cot2A=14 ...(ii)
Squaring (ii) both sides, we get
tan4A+cot4A=194
If sinA-√6 cos A=√7cos A, then cos A+√6sin A is equal to
- (a)
√6 sinA
- (b)
√7sin A
- (c)
√6 cos A
- (d)
√7 cos A
sin A - √6 cos A=√7
⇒ sin A=(√7+√6)cos A
⇒ √7 sin A - √6 sin A =(7-6) cos A
⇒ √7 sin A=cos A +√6 sin A
If \({sin^4\theta\over a}+{cos^4\theta\over b}={1\over a+b},\) then \({sin^{12}\over a^5}+{cos^{12}\theta\over b^5}=\)
- (a)
1/(a+b)5
- (b)
1/(a-b)5
- (c)
1/(a2+b2)5
- (d)
1/(a2+b2)3
If \({sin A\over sin B}=m\ and {cos A\over cos B}=n\) then find the values of tan B; n2 < 1< m2
- (a)
n2
- (b)
\(\pm\sqrt{1-n^2\over m^2-1}\)
- (c)
n2/(m2-1)
- (d)
m2
If \({sinx\over cosx}\times{secx\over cosecx}\times{tanx\over cotx},\) where \(x\in\left(0, {\pi\over2}\right),\) then the value of x is equal
- (a)
π/4
- (b)
π/3
- (c)
π/2
- (d)
π
\({sinx\over cosx}\times{secx\over cosecx}\times{tanx\over cotx}=9⇒tan^4x=9\)
\(⇒tan^2x=3⇒tanx\pm\sqrt3⇒x={\pi\over3}\in\left(0,{\pi\over 2}\right)\)
If tan(A-B)=1, sec(A+B)=\(2\over\sqrt3\) the smallest positive value of B is
- (a)
\(25π\over24\)
- (b)
\(19π\over24\)
- (c)
\(13π\over24\)
- (d)
\(7π\over24\)
tan(A-B)=450=1⇒A-B=450 or 2250
sec(A+B)=\(2\over \sqrt3\)⇒A+B=300 or 330
\(A+B=330^0=2π-{π\over6}={11π\over6}\) ...(i)
and \(A-B=225^0={5π\over 4}\) ..(ii)
Solving (i) and (ii), we get
\(2B={11π\over 6}-{5π\over 4}⇒2B={7π\over 12}⇒B={
7π\over24}\)
Find the value of sin 200sin400sin800
- (a)
\(\sqrt3\over8\)
- (b)
\(\sqrt3\over2\)
- (c)
\(\sqrt3\over4\)
- (d)
\(\sqrt3\)
We have, sin200sin400sin800
=\(1\over2\)[2sin200 .sin400]sin800
=\(1\over2\)[cos(200 -400)-cos(200 + 400)].sin800
=\(1\over2\)[cos(-200)-cos600]sin800
=\({1\over2}\times{1\over2}\left[2\left(cos20^0-{1\over2}\right).sin80^0\right]\)
=\(1\over4\)[2cos00 x sin800 -sin800]
=\(1\over4\)[sin(200 + 800)-sin(200 - 800)-sin800]
=\(1\over4\)[sin1000 +sin600 -sin800]
=\(1\over4\)[sin(1800-800)+sin600 -sin800]
\(={1\over4}[sin80^0+sin60^0-sin80^0]={1\over4}\times{\sqrt{3}\over2}={\sqrt3\over8}\)
If A+B+C=1800, then \(sin^2{A\over2}+sin^2{B\over 2}+sin^2{C\over2}=\)
- (a)
\(1-2cos{A\over 2}cos{B\over2}cos{C\over2}\)
- (b)
\(1-2sin{A\over 2}sin{B\over2}sin{C\over2}\)
- (c)
\(1-4sin{A\over 2}sin{B\over2}sin{C\over2}\)
- (d)
\(1-4cos{A\over 2}cos{B\over2}cos{C\over2}\)
L.H.S.=\(sin^2{A\over 2}+sin^2{B\over2}+sin^2{C\over2}\)
\(={1-cosA\over2}+{1-cosB\over2}+{1-cosC\over2}\)
\(={3-(cosA+cosB+cosC)\over 2}={3-S\over2}\)
where S=cosA+cosB+cosC=(cosA+cosB)+cosC
\(=2cos\left(A+B\over 2\right)cos\left(A-B\over 2\right)+cos\left(2.{C\over 2}\right)\)
\(=2cos\left(90^0{C\over2}\right)cos\left(A-B\over 2\right)+cos\left(2.{C\over 2}\right)\)
\(=2sin{C\over2}cos\left(A-B\over 2\right)+1-2sin^2{C\over2}\)
\(=1+2sin{C\over2}\left\{cos\left(A-B\over 2\right)-2sin{C\over2} \right\}\)
\(=1+2sin{C\over2}\left\{cos\left(A-B\over 2\right)-sin\left(90^0-{A+B\over2}\right) \right\}\)
\(=1+2sin{C\over2}\left\{cos\left(A-B\over 2\right)-cos\left({A+B\over2}\right) \right\}\)
\(=1+2sin{C\over2}\left\{-2sin{A\over2}sin\left\{-{B\over2}\right\}\right\}\)
\(=1+4sin{A\over2}sin{B\over2}sin{C\over2}\)
Then from (i), we get
L.H.S= \(3-\left(1+4sin{A\over2}sin{B\over2}sin{C\over2}\right)\over2\)
\(=1-2sin{A\over2}sin{B\over2}sin{C\over2}\)
The value of \(cos{2\pi\over7}+cos{4\pi\over7}+cos{6\pi\over7}=\)
- (a)
1/2
- (b)
1
- (c)
0
- (d)
-1/2
We have \(cos{2\pi\over7}+cos{4\pi\over7}+cos{6\pi\over7}=\)
=cos2θ+cos4θ+cos6θ, where \(θ={\pi\over7}\)
\(={1\over 2sinθ}\){2cos2θsinθ+2cos4θ+sinθ+2cos6θsinθ}
\(={1\over2sinθ}\){(sin3θ-sinθ+(sin5θ-sin3θ)+(sin7θ-sin5θ)}
\(={1\over2sinθ}(sinθ7θ-sinθ)={1\over2sinθ}(-sinθ)=-{1\over 2}\)
Solve the equation tan(x+a)tan(x+b)+tan(x+c)+tan(x+c)tan(x+a)=1
- (a)
nπ n ∈I
- (b)
\({{n\pi\over 3}}+{\pi\over6}-\left(a+b+c\over3\right), n\in I\)
- (c)
\({n\pi\over 3}-{\pi\over 6}, n\in I\)
- (d)
None of these
Given equation is
tan(x+a)tan(x+b)+tan(x+c)+tan(x+c)tan(x+a)=1
⇒ tan(x+b){tan(x+a)+tan(x+c)}=1-tan(x+c)tan(x+a)
\(⇒tan(x+b)\left\{tan(x+a)+tan(x+c)\over 1-tan(x+a)tan(x+c)\right\}=1\)
⇒ tan(x+a+x+c)= \({1\over tan(x+b)}=cos(x+b)\)
\(⇒tan(2x+x+c)=tan\left({\pi\over2}-(x+b)\right)\)
\(⇒2x+a+c=n\pi+{\pi\over2}-(x+b), n\in I\)
\(⇒3x=n\pi+{\pi\over2}-(a+b+c), n\in I\)
\(⇒ x={n\pi\over2}+{\pi\over 6}-\left(a+b+c\over 3\right), n\in I\)
If 8cos2θ+8sec2θ=65, 0<θ <π/2 then the value of 4cos4θ is equal
- (a)
- 33/2
- (b)
-31/32
- (c)
-31/32
- (d)
-33/32
8cos2θ+8sec2θ=65, 0<θ<π/2
⇒ 8cos22θ+8=65 cos2θ
⇒ 8cos2θ-65 cos2θ+8=0
⇒ (cos2θ-8)(8cos2θ-1)=0
⇒ cos2θ=1/8, cos2θ=8
(cos2θ ≠8 as cos2θ ∈[-1,1])
So, cos2θ=1/8. Now 4cos4θ=4(2cos22θ-1)
=4(2.(1/64)-1)=-31/8
Find the general solution for the equation cos4x==cos2x
- (a)
\({n\pi\over2}n\in Z\)
- (b)
nπ, n∈Z
- (c)
\({nπ\over3},n\in Z\)
- (d)
Both (b) and (c)
We have cos4x=cos2x
∴ the general solution is 4x=2nπ±2x
⇒ 4x=2nπ+2x or 4x=2nπ-2x
⇒ 4x-x2x=2nπ or 4x=2nπ-2x
⇒ x=nπ or x=\(1\over3\) nπ, where n∈Z
x=nπ or x=41/3 where n∈Z
The value of cos10cos20 cos 30...cos1790 is
- (a)
\(1\over\sqrt2\)
- (b)
0
- (c)
1
- (d)
-1
cos10cos20cos30...cos900....cos1790
=cos10cos20cos30....0....cos1790=0
If \(tanα={m\over m+1}, tan\beta={1\over2m+1}\) then α+β is equal to
- (a)
π/2
- (b)
π/3
- (c)
π/6
- (d)
π/4
Given that \(tanα={m\over m+1}, tan\beta={1\over2m+1}\)
\(∴tan(\alpha+\beta)={tanα+tan\beta\over1-tanαtan\beta}={{m\over m+1}+{1\over2m+1}\over 1-{m\over m+1}\times{1\over2m+1}}\)
\(={2m^2+m+m+1\over (m+1)(2m+1)-m}={2m^2+2m+1\over 2m^2+2m+1}=1\)
\(⇒ tan(α+β)=1\Rightarrow \alpha+\beta={\pi\over 4}\)
If \(\alpha\) + \(\beta\) = \(\frac { \pi }{ 4 } \), then the value of (1 + tan \(\alpha\) ) (1 + tan \(\beta\)) is
- (a)
1
- (b)
2
- (c)
-2
- (d)
Not defined
Given , \(\alpha\) + \(\beta\) = \(\frac { \pi }{ 4 } \) \(\Rightarrow\) tan (\(\alpha\) + \(\beta\) ) = tan \(\frac { \pi }{ 4 } \)
\(\Rightarrow\) \(\frac { tan\alpha +tan\beta }{ 1-tan\alpha tan\beta } =1\)
\(\Rightarrow\) tan\(\alpha\) + tan = 1 - tan\(\alpha\) tan\(\beta\)
\(\Rightarrow\) tan\(\alpha\) + tan + tan\(\alpha\) tan\(\beta\) + 1 = 1 + 1
\(\Rightarrow\) (1 + tan \(\alpha\)) + tan\(\beta\) (1 + tan\(\beta\)) = 2
\(\Rightarrow\) (1+ tan\(\alpha\) ) (1 + tan\(\beta\)) = 2
if tan \(\alpha\) = \(\frac { 1 }{ 7 } \) , tan \(\beta\) = \(\frac { 1 }{ 3 } \), then cos 2\(\alpha\) is equal to
- (a)
sin2\(\beta\)
- (b)
sin4\(\beta\)
- (c)
sin3\(\beta\)
- (d)
cos2\(\beta\)
we have, tan \(\alpha\) = \(\frac { 1 }{ 7 } \), tan\(\beta\) = \(\frac { 1 }{ 3 } \)
Now cos2\(\alpha\) = \(\frac { 1-{ tan }^{ 2 }\alpha }{ 1+{ tan }^{ 2 }\alpha } =\frac { 1-\frac { 1 }{ 49 } }{ 1+\frac { 1 }{ 49 } } =\frac { 48 }{ 50 } ......(i)\)
Now sin2\(\beta\) = \(\frac { 2tan\beta }{ 1+{ tan }^{ 2 }\beta } =\frac { 2\times \frac { 1 }{ 3 } }{ 1+\frac { 1 }{ 9 } } =\frac { 2 }{ 3 } \times \frac { 9 }{ 10 } =\frac { 3 }{ 4 } ...(ii)\)
Also , sin4\(\beta\) = 2sin\(\beta\) cos2\(\beta\) = \(2.\frac { 3 }{ 5 } \left( \frac { 1-{ tan }^{ 2 }\beta }{ 1+{ tan }^{ 2 }\beta } \right) \) (using (ii))
= \(\frac { 6 }{ 5 } \left( \frac { 1-\frac { 1 }{ 9 } }{ 1+\frac { 1 }{ 9 } } \right) =\frac { 6 }{ 5 } \times \frac { 8 }{ 9 } \times \frac { 9 }{ 10 } =\frac { 48 }{ 50 } \)
Hence, cos2\(\alpha\) = sin 4\(\beta\) (from (i) and (iii))
if for real values of x, cos \(\theta\) = x + \(\frac { 1 }{ x } \), then
- (a)
\(\theta\) is an acute angle
- (b)
\(\theta\) is right angle
- (c)
\(\theta\) is an obtuse angle
- (d)
No value of \(\theta\) os possible
We have , cos \(\theta\) = x + \(\frac { 1 }{ x } \)
Since x + \(\frac { 1 }{ x } \ge 2\sqrt { x.\frac { 1 }{ x } } \) (using property A.M \(\ge\) G.M)
\(\Rightarrow\) \(x+\frac { 1 }{ x } \ge 2\Rightarrow cos\theta \ge 2\)
But - 1 \(\le\) cos \(\theta\) \(\le\) 1
\(\therefore\) From (i) and (ii) , we conclude that no value of \(\theta\) is possible
let \(\alpha\) be a real number lying between 0 and \(\frac { \pi }{ 2 } \) and n be a positive integer.
Statement -I: tana + 2tan2\(\alpha\) + 22tan22\(\alpha\) + .... + 2n-1 tan2n-1 \(\alpha\) + 2n cot 2n\(\alpha\) = cot \(\alpha\)
Statement-II : cot\(\alpha\) - tan\(\alpha\) =2cot2\(\alpha\)
- (a)
If both statement-I and statement -II are trur and statement -II is the correct explantion of statement -I
- (b)
If both statement -I and statement II are true but statment-II is not the correct explanation of statement-I
- (c)
If statement -I is true but statement -II is false
- (d)
if statement -I is false and statement -II is true
Now, cot \(\alpha\) - tan \(\alpha\) = \(\frac { 1 }{ tan\alpha } -tan\alpha =\frac { 1-{ tan }^{ 2 }\alpha }{ tan\alpha } \)
= \(2\left( \frac { 1-{ tan }^{ 2 }\alpha }{ 2tan\alpha } \right) =2cot2\alpha \)
Fom here, we get tan\(\alpha\) = cot\(\alpha\) - 2cot2\(\alpha\)
Making repated use of this identity , we shall obtain
tan\(\alpha\) + 2tan2\(\alpha\) + 22tan22\(\alpha\) + ... + 2n-1 tan2n-1 \(\alpha\) + 2ncot2n\(\alpha\)
= (cot\(\alpha\) - 2cot2\(\alpha\)) + 2(cot2\(\alpha\) - 2cot22\(\alpha\)) + 22 (cot22\(\alpha\) - 2cot23\(\alpha\)) + ...+2n-1 \(\alpha\) - 2cot2n\(\alpha\) + 2ncot2n\(\alpha\)
=cot \(\alpha\)
Statement -I: The solution of the equation
tan \(\theta\) + tan \(\left( \theta +\frac { \pi }{ 3 } \right) +tan\left( \theta +\frac { 2\pi }{ 3 } \right) \) = 3 is \(\theta\) = \(\frac { n\pi }{ 3 } +\frac { \pi }{ 12 } ,n\epsilon I\)
Statement II : if tan \(\theta\) = tan \(\alpha\), then \(\theta\) = n\(\pi\) + \(\alpha\), \(n\epsilon I\)
- (a)
If both statement-I and statement -II are trur and statement -II is the correct explantion of statement -I
- (b)
If both statement -I and statement II are true but statment-II is not the correct explanation of statement-I
- (c)
If statement -I is true but statement -II is false
- (d)
if statement -I is false and statement -II is true
Given equation is
\(tan\theta +tan\left( \theta +\frac { \pi }{ 3 } \right) +tan\left( \theta +\frac { 2\pi }{ 3 } \right) \) = 3
\(\Rightarrow\) \(tan\theta +\frac { tan\theta +\sqrt { 3 } }{ 1-\sqrt { 3 } tan\theta } +\frac { tan\theta -\sqrt { 3 } }{ 1+\sqrt { 3 } tan\theta } =3\)
\(\Rightarrow\) \(tan\theta +\frac { \left( tan\theta +\sqrt { 3 } \right) (1+\sqrt { 3 } tan\theta )+(tan\theta -\sqrt { 3 } )x\left( 1-\sqrt { 3 } tan\theta \right) }{ \left( 1-\sqrt { 3 } tan\theta \right) \left( 1+\sqrt { 3 } tan\theta \right) } =3\)
\(\Rightarrow\) \(tan\theta +\frac { 8tan\theta }{ 1-3{ tan }^{ 2 }\theta } =3\Rightarrow \frac { tan\theta -3{ tan }^{ 3 }\theta +8tan\theta }{ 1-3{ tan }^{ 2 }\theta } =3\)
\(\Rightarrow\) \(\frac { 3\left( 3tan\theta -{ tan }^{ 3 }\theta \right) }{ 1-3{ tan }^{ 2 }\theta } =3\Rightarrow 3tan3\theta =3\Rightarrow tan3\theta =1\)
\(\Rightarrow\) \(tan3\theta =tan\frac { \pi }{ 4 } \Rightarrow 3\theta -n\pi +\frac { \pi }{ 4 } ,n\epsilon I\)
\(\Rightarrow\) \(\theta =\frac { n\pi }{ 3 } +\frac { \pi }{ 12 } ,n\epsilon I\)
Statement-I : If a, b , \(\theta\) are real numbers , then minimum and maximum values of asin\(\theta\) + bcos\(\theta\) are respectively - \(\sqrt { { a }^{ 2 }+{ b }^{ 2 } } \) and \(\sqrt { { a }^{ 2 }+{ b }^{ 2 } } \).
Statement II : the equation asin\(\theta\) + bcos\(\theta\) = c (a, b,c \(\epsilon \), R)has a solution only it c2 > a2 + b2
- (a)
If both statement-I and statement -II are trur and statement -II is the correct explantion of statement -I
- (b)
If both statement -I and statement II are true but statment-II is not the correct explanation of statement-I
- (c)
If statement -I is true but statement -II is false
- (d)
if statement -I is false and statement -II is true
Statement-I is a standard result.
For the equation asin \(\theta\) + bcos \(\theta\) = c to have a solution we must have
- \(-\sqrt { { a }^{ 2 }+{ b }^{ 2 } } \le c\le \sqrt { { a }^{ 2 }+{ b }^{ 2 } } \)
\(\Leftrightarrow \) |c| \(\le\) \(\sqrt { { a }^{ 2 }+{ b }^{ 2 } } \) \(\Leftrightarrow \) c2 \(\le\) a2 + b2
Fill in the blanks.
(i) The value of the expression
\(3\left[sin^4\left({3\pi\over2}-\alpha\right)+sin^4(3\pi+\alpha)\right]-2\left[sin^6\left({\pi\over2}+\alpha\right)+sin^6(5\pi-\alpha)\right]\)is P
(ii) The value of the expression
\(coa^4{\pi\over8}+cos^4{3\pi\over 8}+cos^4{5\pi\over 8}+cos^4{7\pi\over8}\)is Q
(iii) If θ lies in the first quadrant and cos θ = \(8\over 17\)
then the value of
cos(30° + θ) + cos(45° - θ) + cos(l20° -θ) is R
- (a)
P Q R 2 \(1\over 2\) \({13\over 17}\left({\sqrt3-1\over 2}+{1\over \sqrt3}\right)\) - (b)
P Q R 1 \(3\over 2\) \({23\over 17}\left({\sqrt3-1\over 2}+{1\over \sqrt2}\right)\) - (c)
P Q R 3 \(3\over 2\) \({23\over 17}\left({\sqrt3-1\over 2}+{1\over \sqrt3}\right)\) - (d)
P Q R 1 \(1\over 2\) \({13\over 17}\left({\sqrt3-1\over 2}+{1\over \sqrt2}\right)\)
If \(\left( \frac { sin\theta }{ sin\phi } \right) ^{ 2 }=\frac { tan\theta }{ tan\theta } =3,\) then the value of \(\theta \) and \(\phi \) are \(\theta =n\pi \pm \frac { \pi }{ A } ,\phi =n\pi \pm \frac { \pi }{ B } \) Find A+B.
- (a)
8
- (b)
9
- (c)
10
- (d)
11
\(\left( \frac { sin\theta }{ sin\theta } \right) ^{ 2 }=\frac { tan\theta }{ tan\phi } \)
\(\Rightarrow sin\theta .cos\theta =sin\phi cos\phi \)
\(\Rightarrow sin2\theta =sin2\phi \)
\(2\theta =\pi -2\phi \Rightarrow \theta =\frac { \pi }{ 2 } =\phi \)
\(But\frac { tan\theta }{ tan\phi } =3\Rightarrow \frac { tan\theta }{ cot\theta } =3\Rightarrow tan^{ 2 }\theta =3\)
\(\Rightarrow \theta =n\pi \pm \frac { \pi }{ 3 } ,so\quad that\quad \phi =n\pi \pm \frac { \pi }{ 6 } \)
A = 3, B = 6
1654060945.png)
1654060942.png)
1653987551.png)
1653987086.png)
1582095229.jpg)
