Practice Start Test

Trigonometric Ratio and Identities

JEE Mathematics · 51 questions · Page 1 of 6 · Click an option or "Show Solution" to reveal answer

Q1
If ∑r=113{1sin⁡(π4+(r−1)π6)sin⁡(π4+rπ6)}=a3+b,a,b∈Z\sum\limits_{r=1}^{13}\left\{\dfrac{1}{\sin \left(\dfrac{\pi}{4}+(r-1) \dfrac{\pi}{6}\right) \sin \left(\dfrac{\pi}{4}+\dfrac{r \pi}{6}\right)}\right\}=a \sqrt{3}+b, a, b \in Z, then a2+b2a^2+b^2 is equal to :
A 10
B 4
C 8
D 2
Correct Answer
Option C
Solution
1sin⁡π6∑r=113sin⁡[(π4+rπ6)−(π4)−(r−1)π6]sin⁡(π4+(r−1)π6)sin⁡(π4+rπ6)1sin⁡π6∑r=113(cot⁡(π4+(r−1)π6)−cot⁡(π4+rπ6))=23−2=α3+b So a2+b2=8\begin{aligned} &\begin{aligned} & \frac{1}{\sin \frac{\pi}{6}} \sum_{r=1}^{13} \frac{\sin \left[\left(\frac{\pi}{4}+\frac{\mathrm{r} \pi}{6}\right)-\left(\frac{\pi}{4}\right)-(\mathrm{r}-1) \frac{\pi}{6}\right]}{\sin \left(\frac{\pi}{4}+(\mathrm{r}-1) \frac{\pi}{6}\right) \sin \left(\frac{\pi}{4}+\frac{\mathrm{r} \pi}{6}\right)} \\ & \frac{1}{\sin \frac{\pi}{6}} \sum_{\mathrm{r}=1}^{13}\left(\cot \left(\frac{\pi}{4}+(\mathrm{r}-1) \frac{\pi}{6}\right)-\cot \left(\frac{\pi}{4}+\frac{\mathrm{r} \pi}{6}\right)\right) \\ & =2 \sqrt{3}-2=\alpha \sqrt{3}+\mathrm{b} \end{aligned}\\ &\text{ So } \mathrm{a}^2+\mathrm{b}^2=8 \end{aligned}
Q2
If x=∑n=0∞(−1)ntan⁡2nθx = \sum\limits_{n = 0}^\infty {{{\left( { - 1} \right)}^n}{{\tan }^{2n}}\theta } and y=∑n=0∞cos⁡2nθy = \sum\limits_{n = 0}^\infty {{{\cos }^{2n}}\theta } for 0 < θ\theta < π4{\pi \over 4}, then :
A x(1 + y) = 1
B y(1 – x) = 1
C y(1 + x) = 1
D x(1 – y) = 1
Correct Answer
Option B
Solution
x=∑n=0∞(−1)ntan⁡2nθx = \sum\limits_{n = 0}^\infty {{{\left( { - 1} \right)}^n}{{\tan }^{2n}}\theta }

= 1 – tan2θ\theta + tan2 4θ\theta + ... =

11+tan⁡2θ{1 \over {1 + {{\tan }^2}\theta }}

= cos2 θ\theta ....(1)

y=∑n=0∞cos⁡2nθy = \sum\limits_{n = 0}^\infty {{{\cos }^{2n}}\theta }

= 1 + cos2 θ\theta + cos4 θ\theta + cos6 θ\theta + .... =

11−cos⁡2θ{1 \over {1 - {{\cos }^2}\theta }}

=

1sin⁡2θ{1 \over {{{\sin }^2}\theta }}

⇒\Rightarrow sin2 θ\theta =

1y{1 \over y}

...(2) Adding (1) and (2), we get, x +

1y{1 \over y}

= sin2 θ\theta + cos2 θ\theta ⇒\Rightarrow x +

1y{1 \over y}

= 1 ⇒\Rightarrow y(1 – x) = 1

Q3
The value of (sin⁡70∘)(cot⁡10∘cot⁡70∘−1)\left(\sin 70^{\circ}\right)\left(\cot 10^{\circ} \cot 70^{\circ}-1\right) is
A 0
B 2/3
C 1
D 3/2
Correct Answer
Option C
Solution
sin⁡70∘(cot⁡10∘cot⁡70∘−1)⇒cos⁡(80∘)sin⁡10=1\begin{aligned} & \sin 70^{\circ}\left(\cot 10^{\circ} \cot 70^{\circ}-1\right) \\ & \Rightarrow \frac{\cos \left(80^{\circ}\right)}{\sin 10}=1 \end{aligned}
Q4
If cotα\alpha = 1 and secβ\beta = −53 - {5 \over 3}, where $$\pi
A −17 - {1 \over 7} and IVth quadrant
B 7 and Ist quadrant
C −-7 and IVth quadrant
D 17 {1 \over 7} and Ist quadrant
Correct Answer
Option A
Solution

∵cot⁡α=1,α∈(π,3π2)\because \cot \alpha=1, \quad \alpha \in\left(\pi, \dfrac{3 \pi}{2}\right) then tan⁡α=1\tan \alpha=1 and sec⁡β=−53,β∈(π2,π)\sec \beta=-\dfrac{5}{3}, \quad \beta \in\left(\dfrac{\pi}{2}, \pi\right) then tan⁡β=−43\tan \beta=-\dfrac{4}{3} ∴tan⁡(α+β)=tan⁡α+tan⁡β1−tan⁡α⋅tan⁡β\therefore \tan (\alpha+\beta)=\dfrac{\tan \alpha+\tan \beta}{1-\tan \alpha \cdot \tan \beta}

=1−431+43=−17\begin{aligned} &=\frac{1-\frac{4}{3}}{1+\frac{4}{3}} \\\\ &=-\frac{1}{7} \end{aligned}
α+β∈(3π2,2π) i.e. fourth quadrant \alpha+\beta \in\left(\frac{3 \pi}{2}, 2 \pi\right) \text{ i.e. fourth quadrant }
Q5
The value of sin 10º sin30º sin50º sin70º is :-
A 136{1 \over {36}}
B 116{1 \over {16}}
C 132{1 \over {32}}
D 118{1 \over {18}}
Correct Answer
Option B
Solution

sin 10º sin30º sin50º sin70º = sin30º sin50º sin 10º sin70º =

12{1 \over 2}

[ sin50º sin 10º sin70º ] =

12{1 \over 2}

[ sin(60º - 10º) sin 10º sin(60º + 10º) ] =

12{1 \over 2}

[

14sin⁡{1 \over 4}\sin

3(10º) ] =

12{1 \over 2}

[

14×12{1 \over 4} \times {1 \over 2}

] =

116{1 \over {16}}

Note :

sin⁡(60∘−A)sin⁡Asin⁡(60∘−A)=14sin⁡3A\sin \left( {60^\circ - A} \right)\sin A\sin \left( {60^\circ - A} \right) = {1 \over 4}\sin 3A
Q6
The equation y = sinx sin (x + 2) – sin2 (x + 1) represents a straight line lying in :
A first, second and fourth quadrants
B first, third and fourth quadrants
C second and third quadrants only
D third and fourth quadrants only
Correct Answer
Option D
Solution

y = sinx.sin(x+2) - sin2(x+1)

⇒12{2sin⁡(x+2)sin⁡x−2sin⁡2(x+1)}\Rightarrow {1 \over 2}\left\{ {2\sin \left( {x + 2} \right)\sin x - 2{{\sin }^2}(x + 1)} \right\}
⇒12{cos⁡2−cos⁡(2x+2)+cos(2x+2)−1}=−sin⁡21<0\Rightarrow {1 \over 2}\left\{ {\cos 2 - \cos (2x + 2) + cos(2x + 2) - 1} \right\} = - {\sin ^2}1 < 0

Hence the line passes through III and IV quadrant

Q7
The value of cos210° – cos10°cos50° + cos250° is
A 32+cos⁡20o{3 \over 2} + \cos {20^o}
B 34{3 \over 4}
C 32(1+cos⁡20o){3 \over 2}(1 + \cos {20^o})
D 32{3 \over 2}
Correct Answer
Option B
Solution

cos210° – cos10°cos50° + cos250° =

12{1 \over 2}

[ 2cos210° – 2cos10°cos50° + 2cos250°] =

12{1 \over 2}

[ 1 + cos20° - cos60° - cos40° + 1 + cos100°] =

12{1 \over 2}

[ 2 -

12{1 \over 2}

+ cos20° + cos100° - cos40°] =

12{1 \over 2}

[

32{3 \over 2}

+ 2cos60°cos40° - cos40°] =

12{1 \over 2}

[

32{3 \over 2}

+ cos40° - cos40°] =

34{3 \over 4}
Q8
If cos(α\alpha + β\beta ) = 3/5 ,sin ( α\alpha - β\beta ) = 5/13 and 0 < α,β\alpha , \beta < π4\pi \over 4, then tan(2α\alpha ) is equal to :
A 21/16
B 63/52
C 33/52
D 63/16
Correct Answer
Option D
Solution

Given

0<α<π40 < \alpha < {\pi \over 4}

and

0<β<π40 < \beta < {\pi \over 4}

∴\therefore

0>−β>−π40 > - \beta > - {\pi \over 4}

∴\therefore

0<α+β<π20 < \alpha + \beta < {\pi \over 2}

and

−π4<α−β<π4- {\pi \over 4} < \alpha - \beta < {\pi \over 4}

As cos(α\alpha + β\beta) = 3/5 so

tan⁡(α+β)=43{\tan \left( {\alpha + \beta } \right) = {4 \over 3}}

As sin( α\alpha - β\beta) = 5/13 so

tan⁡(α−β)=512{\tan \left( {\alpha - \beta } \right) = {5 \over {12}}}

Now tan(2α\alpha) = tan(α\alpha + β\beta + α\alpha - β\beta) =

tan⁡(α+β)+tan⁡(α−β)1−tan⁡(α+β)tan⁡(α−β){{\tan \left( {\alpha + \beta } \right) + \tan \left( {\alpha - \beta } \right)} \over {1 - \tan \left( {\alpha + \beta } \right)\tan \left( {\alpha - \beta } \right)}}

=

43+5121−43×512{{{4 \over 3} + {5 \over {12}}} \over {1 - {4 \over 3} \times {5 \over {12}}}}

=

6316{{63} \over {16}}
Q9
The maximum value of 3cosθ\theta + 5sin (θ−π6)\left( {\theta - {\pi \over 6}} \right) for any real value of θ\theta is :
A 34\sqrt {34}
B 31\sqrt {31}
C 19\sqrt {19}
D 792{{\sqrt {79} } \over 2}
Correct Answer
Option C
Solution

y = 3cosθ\theta + 5

(sin⁡θ32−cos⁡θ12)\left( {\sin \theta {{\sqrt 3 } \over 2} - \cos \theta {1 \over 2}} \right)
532{{5\sqrt 3 } \over 2}

sinθ\theta +

12{1 \over 2}

cosθ\theta ymax =

754+14\sqrt {{{75} \over 4} + {1 \over 4}}

=

19\sqrt {19}
Q10
If e(cos⁡2x+cos⁡4x+cos⁡6x+...∞)log⁡e2{e^{\left( {{{\cos }^2}x + {{\cos }^4}x + {{\cos }^6}x + ...\infty } \right){{\log }_e}2}} satisfies the equation t2 - 9t + 8 = 0, then the value of 2sin⁡xsin⁡x+3cos⁡x(0<x<π2){{2\sin x} \over {\sin x + \sqrt 3 \cos x}}\left( {0 < x < {\pi \over 2}} \right) is :
A 3\sqrt 3
B 32{3 \over 2}
C 23\sqrt 3
D 12{1 \over 2}
Correct Answer
Option D
Solution
e(cos⁡2x+cos⁡4x+...........∞)ln⁡2=2cos⁡2x+cos⁡4x+...........∞{e^{({{\cos }^2}x + {{\cos }^4}x + ...........\infty )\ln 2}} = {2^{{{\cos }^2}x + {{\cos }^4}x + ...........\infty }}

=

2cos⁡2x1−cos⁡2x{2^{{{{{\cos }^2}x} \over {1 - {{\cos }^2}x}}}}
=2cot⁡2x= {2^{{{\cot }^2}x}}

Given,

t2−9t+8=0⇒t=1,8{t^2} - 9t + 8 = 0 \Rightarrow t = 1,8
⇒2cot⁡2x=1,8⇒cot2x=0,3\Rightarrow {2^{{{\cot }^2}x}} = 1,8 \Rightarrow co{t^2}x = 0,3
0<x<π2⇒cot⁡x=30 < x < {\pi \over 2} \Rightarrow \cot x = \sqrt 3

∴\therefore

2sin⁡xsin⁡x+3cos⁡x=21+3cot⁡x=24=12{{2\sin x} \over {\sin x + \sqrt 3 \cos x}} = {2 \over {1 + \sqrt 3 \cot x}} = {2 \over 4} = {1 \over 2}
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