Mathematics

Trigonometric Identities

82 Questions

Trigonometric identities focus on solving equations using tangent, sine, and cosine properties. Questions involve half angle formulas and slope calculations. This topic is a staple in the quantitative aptitude section of major competitive examinations.

Tangent propertiesAngle formulasTrigonometric equationsHalf angle identitiesSlope calculations

Trigonometric Identities Questions

Multiple choice maths theorems on triangles theorem of remote interior angles of a triangle use of properties of parallel lines angle sum property of a triangle

In $\Delta ABC$, if $\angle A+\angle B=90^{\circ}$, cot $B=\dfrac{3}{4}$, then the value of tan A is :

  1. $\dfrac{4}{5}$
  2. $\dfrac{3}{4}$
  3. $\dfrac{4}{3}$
  4. $\dfrac{3}{5}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\angle A+\angle B={ 90 }^{ \circ  }\ \angle B={ 90 }^{ \circ  }-\angle A$

$\cot { B } =\dfrac { 3 }{ 4 } \ \cot { \left( { 90 }^{ \circ  }-\angle A \right)  } =\dfrac { 3 }{ 4 } \ \tan { A } =\dfrac { 3 }{ 4 } $

Multiple choice maths theorems on triangles theorem of remote interior angles of a triangle use of properties of parallel lines angle sum property of a triangle

In $\Delta ABC$. If $x=\tan\left(\dfrac{B-C}{2}\right)\tan\dfrac{A}{2}, y=\tan\left(\dfrac{C-A}{2}\right)\tan\dfrac{B}{2}, z=\tan\left(\dfrac{A-B}{2}\right)\tan\dfrac{C}{2}$, then $x+y+z$ (in terms of $x,y,z$ only) is 

  1. $xyz$
  2. $2xyz$
  3. $-xyz$
  4. $\dfrac{1}{2}xyz$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In $ \triangle ABC$


$\tan\left(\dfrac{B-C}{2}\right)=\dfrac{b-c}{b+c}\cot\dfrac{A}{2}$


$\implies x=\dfrac{b-c}{b+c}\implies\dfrac{-c}{b}$


Similarly $y=\dfrac{-a}{c},z=\dfrac{-b}{a}$

These on adding gives $\dfrac{-(ac^2+ba^2+cb^2)}{(abc)^2}$

Multiple choice maths differencial calculus - differenciability and methods of differnciation differentiation by substitution methods of differentiation derivative of a function

If $y=sec(tan^{-1}x)$, then $\displaystyle\frac{dy}{dx}$ is.

  1. $\displaystyle\frac{x}{\sqrt{1+x^2}}$
  2. $\displaystyle\frac{-x}{\sqrt{1+x^2}}$
  3. $\displaystyle\frac{x}{\sqrt{1-x^2}}$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given, $y=sec(\tan^{-1}x)$
On differentiating w.r.t. $x,$ we get
$\displaystyle\frac{dy}{dx}=sec(\tan^{-1}x)\cdot \tan(\tan^{-1}x)\frac{1}{1+x^2}$
$=\displaystyle\frac{x}{1+x^2}\sqrt{1+x^2}$
$[\because \tan^{-1} x=sec^{-1}(\sqrt{1+x^2})]$
$=\displaystyle\frac{x}{\sqrt{1+x^2}}$

Multiple choice maths differencial calculus - differenciability and methods of differnciation differentiation by substitution methods of differentiation derivative of a function

If $y=(tan \, x)^{(tan\, x)^{tan\,x}}, $ then at $x=\dfrac{\pi}{4}, \dfrac{dy}{dx}$ is equal to 

  1. 0

  2. 3

  3. 2

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$y={ \left( tanx \right)  }^{ { \left( tanx \right)  }^{ \left( tanx \right)  } }$
We have to find $\dfrac { dy }{ dx } $ at $x=\dfrac { \Pi  }{ 4 } $.
Now consider:
$y={ f\left( x \right)  }^{ g\left( x \right)  }$
$ln\left( y \right) =g\left( x \right) lnf\left( x \right) $
$\Rightarrow \dfrac { 1 }{ y } \dfrac { dy }{ dx } =g\left( x \right) \dfrac { { f }^{ 1 }\left( x \right)  }{ f\left( x \right)  } +{ g }^{ 1 }\left( x \right) lnf\left( x \right) $
$\Rightarrow \dfrac { dy }{ dx } =y\left[ g\left( x \right) \dfrac { { f }^{ 1 }\left( x \right)  }{ f\left( x \right)  } +{ g }^{ 1 }\left( x \right) lnf\left( x \right)  \right] $
$\Rightarrow \dfrac { dy }{ dx } ={ f\left( x \right)  }^{ g\left( x \right)  }\left[ { g }^{ 1 }\left( x \right) lnf\left( x \right) +g\left( x \right) \dfrac { { f }^{ 1 }\left( x \right)  }{ f\left( x \right)  }  \right] $
So, We have
$y={ \left( tanx \right)  }^{ { \left( tanx \right)  }^{ tanx } }$
Where  $f\left( x \right) =tan\left( x \right) $
${ f }^{ 1 }\left( x \right) ={ \sec }^{ 2 }x$
$g\left( x \right) ={ \left( tanx \right)  }^{ tanx }$
${ g }^{ 1 }\left( x \right) ={ \left( tanx \right)  }^{ tanx\left[ { \sec }^{ 2 }xln\left( tan\left( x \right)  \right) +\dfrac { \tan\left( x \right) { \sec }^{ 2 }x }{ tanx }  \right]  }$
Now, $\tan\left( \dfrac { \Pi  }{ 4 }  \right) =1$ and $\sec\left( \Pi /4 \right) =\sqrt { 2 } $.
$f\left( \Pi /4 \right) =1$
${ f }^{ 1 }\left( \Pi /4 \right) =2$
$g\left( \Pi /4 \right) =1$
${ g }^{ 1 }\left( \Pi /4 \right) =1\left( 2ln\left( 1 \right) +2 \right) =2$
$\dfrac { dy }{ dx } $ at $\Pi /4$ is
$\dfrac { dy }{ dx } ={ f\left( \Pi /4 \right)  }^{ g\left( \Pi /4 \right) \left[ { g }^{ 1 }\left( \Pi /4 \right) ln\left( f\left( \Pi /4 \right)  \right) +g\left( \Pi /4 \right) \dfrac { { f }^{ 1 }\left( \Pi /4 \right)  }{ f\left( \Pi /4 \right)  }  \right]  }$
$\Rightarrow \dfrac { dy }{ dx } =1\left[ 2ln\left( 1 \right) +1\times 2 \right] =2$
$\Rightarrow \dfrac { dy }{ dx } =2$.
Multiple choice mathematics and statistics coordinates, points and lines what is meant by the equation of a straight line or of a curve introduction to slope introduction to straight lines

Find the slope of $\displaystyle x\cot  \alpha -y\tan \alpha =1$

  1. $\displaystyle cosec^{2}\alpha -1$
  2. $\displaystyle \sec ^{2}\alpha -1$
  3. $\displaystyle \tan ^{2}\alpha -1$
  4. $\displaystyle \cot ^{2}\alpha -1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given, $ x \cot \alpha - y \tan \alpha = 1 $
$ =>  y \tan \alpha = x \cot \alpha - 1 $
$ => y = x \dfrac {\cot \alpha}{\tan \alpha}  - \dfrac {1}{\tan \alpha} $

The slope of the equation of line  of the form $ y = mx + c $ is $ m $
So, slope of the given line $ = \dfrac {\cot \alpha}{\tan \alpha} = \cot ^ 2 (\alpha) = cosec^2 (\alpha) - 1 $

Multiple choice mathematics and statistics angle and their measurement angles and sides naming the sides in a right angled triangle understanding ratios

In $\Delta ABC$ if $a=8,b=9,c=10$, then the value of $\dfrac{{\tan C}}{{\sin B}}$ is

  1. $\dfrac{{32}}{9}$
  2. $\dfrac{{24}}{7}$
  3. $\dfrac{{21}}{4}$
  4. $\dfrac{{18}}{5}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In $\triangle ABC,a=8,b=9,c=10$

as we know that
$\dfrac{sin C}{\sin B}=\dfrac{c}{b}=\dfrac{10}{9}$
and also $\cos C=\dfrac{a^2+b^2-c^2}{2 a b}=\dfrac{8^2+9^2-10^2}{2\times 8\times 9}=\dfrac{5}{16}$
$\dfrac{\tan C}{\sin B}=\dfrac{\sin C}{\sin B}\times \dfrac{1}{\cos C}=\dfrac{10}{9}\times \dfrac{16}{5}=\dfrac{32}{9}$

Multiple choice mathematics and statistics angle and their measurement angles and sides naming the sides in a right angled triangle understanding ratios

If $t _1=(\tan x)^{\cot x}, t _2=(\cot x)^{\cot x}, t _3=(\tan x)^{\tan x}, t _4=(\cot x)^{\tan x}, 0 < x < \dfrac{\pi}{4}$, then:

  1. $t _1 < t _2 < t _3 < t _4$
  2. $t _2 > t _4 > t _3 > t _1$
  3. $t _1 > t _4 > t _3 > t _2$
  4. $t _1 > t _2 > t _3 > t _4$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Since $0 < x < \dfrac{\pi}{4}$

$\therefore \tan x < 1$ and $\cot x > 1$ .....$(1)$
$\therefore$ Choose $\tan x=1-k _1$ and $\cot x=1+k _2$, where 

$k _1$ and $k _2$ are very small $+$ve quantities.
$\therefore t _3=(1-k _1)^{1-k _1}, t _1=(1-k _1)^{1+k _2}$  ....$(2)$
$t _4=(1+k _2)^{1-k _1}, t _2=(1+k _2)^{1+k _2}$  .....$(3)$
$\therefore t _4 > t _3$ by $(3)$ and $(2)$, $t _2 > t _4$ by $(2)$ and $(3)$
$\therefore t _2 > t _4 > t _3$. Also $t _3 > t _1$
$\therefore t _2 > t _4 > t _3 > t _1$.

Multiple choice mathematics and statistics angle and their measurement angles and sides naming the sides in a right angled triangle understanding ratios

The points of discontinuity of $\tan{x}$ are

  1. $n\pi ,n\in I$
  2. $2n\pi ,n\in I$
  3. $(2n+1)\cfrac { \pi }{ 2 } ,n\in I$

  4. None of the above
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let $f(x)=\tan {x}$
The points of discontinuity of $f(x)$ are those points where $\tan {x}$ is infinite. This gives
$\tan { x } =\infty $
ie $\tan { x } =\tan { \cfrac { \pi  }{ 2 }  } $
$x=\left( 2n+1 \right) \cfrac { \pi  }{ 2 } ,n\in I$

Multiple choice mathematics and statistics angle and their measurement angles and sides naming the sides in a right angled triangle understanding ratios

If $\tan A = \dfrac {1 - \cos B}{\sin B}$, then the value of $\dfrac {2\tan A}{1 - \tan^{2}A}$ is

  1. $\dfrac {(\tan B)}{2}$
  2. $2\tan B$
  3. $\tan B$
  4. $4\tan B$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given, $\tan A =  \dfrac {1 - \cos B}{\sin B} $


                       $= \dfrac {2\sin^{2}\dfrac {B}{2}}{2\sin \dfrac {B}{2}\cdot \cos \dfrac {B}{2}}$


                       $= \tan \dfrac {B}{2}$

Therefore, $A = \dfrac {B}{2} \Rightarrow 2A = B$

Now $\dfrac {2\tan A}{1 - \tan^{2}A} = \tan 2A = \tan B$

Multiple choice the nth roots of unity complex numbers maths
Let $z$ be any complex number. To factorise the expression of the form ($z^n- 1$), we consider the equation $z^n = 1$. This equation is solved using De moiver's theorem. Let $1, \alpha _1, \alpha _2 .......\alpha _{n-1}$ be the roots of this equation, then $z^n-1=(z-1)(z-\alpha _1)(z-\alpha _2).......(z-\alpha _{n-1})$. This method can be generalised to factorize any expression of the form $z^n-k^n$.
For example, $z^7+1=\displaystyle \Pi _{m=0}^6\left (z-CiS\left (\dfrac {2m\pi}{7}+\dfrac {\pi}{7}\right )\right )$
This can be further simplified as
$z^7+1(z+1)(z^2-2zcos\dfrac {\pi}{7}+1)(z^2-2z cos\dfrac {3\pi}{7}+1)(z^2-2z cos \dfrac {5\pi}{7}+1)$ .......$(i)$
These factorisations are useful in proving different trigonometric identities e.g. in equation $(i)$ if we put $z = i,$ then equation $(i)$ becomes
$(1-i)=(i+1)(-2i cos \dfrac {\pi}{7})(-2i cos \dfrac {3\pi}{7})(-2i cos \dfrac {5\pi}{7})$
i.e, $cos \dfrac {\pi}{7}cos \dfrac {3\pi}{7}cos \dfrac {5\pi}{7}=-\dfrac {1}{8}$


By using the factorisation for $z^5+1$, the value of $4 sin\dfrac {\pi}{10} cos \dfrac {\pi}{5}$ comes out to be :

  1. $4$
  2. $1/4$
  3. $1$
  4. $-1$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$z^{5}+1=0$ Implies 

$z=-1,(cos\dfrac{\pi}{5}+isin\dfrac{\pi}{5}),(cos\dfrac{-\pi}{5}+isin\dfrac{-\pi}{5}),(cos\dfrac{3\pi}{5}+isin\dfrac{3\pi}{5}),(cos\dfrac{-3\pi}{5}+isin\dfrac{-3\pi}{5})$
Now 
$4sin\dfrac{\pi}{10}.cos\dfrac{\pi}{5}$
$=4sin\dfrac{\pi}{10}sin\dfrac{3\pi}{10}$

$=2[cos\dfrac{\pi}{5}-cos\dfrac{2\pi}{5}]$

$=2[cos\dfrac{\pi}{5}+cos\dfrac{3\pi}{5}]$

$=cos\dfrac{\pi}{5}+cos\dfrac{\pi}{5}+cos\dfrac{3\pi}{5}+cos\dfrac{3\pi}{5}$

$=cos\dfrac{\pi}{5}+cos\dfrac{3\pi}{5}+cos\dfrac{-3\pi}{5}+cos\dfrac{-\pi}{5}$

$=\sum Re z _{i}$

$=|z|$
$=1$

Multiple choice maths binomial theorem, sequence and series series introduction to series introduction to sequences and series

If  in traingle ABC $\cos 2B=\dfrac {\cos (A+C)}{\cos (A-C)}$, then 

  1. $\tan A, \tan B, \tan C$ are in $A.P$
  2. $\tan A, \tan B, \tan C$ are in $G.P$
  3. $\tan A, \tan B, \tan C$ are in $H.P$
  4. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given cos(2B) = cos(A+C)/cos(A-C). Using componendo and dividendo, (1-cos(2B))/(1+cos(2B)) = (cos(A-C)-cos(A+C))/(cos(A-C)+cos(A+C)). This simplifies to tan^2(B) = tan(A)tan(C), meaning tan(A), tan(B), tan(C) are in G.P.

Multiple choice physics trigonometrical ratios angles and sides naming the sides in a right angled triangle angle and their measurement

If $E. \ tan(x -
30^{\circ}) = j. \ tan(x+120^{\circ})$, then $\frac{E + J}{E-J} =$

  1. $\ sin 2x$
  2. $2 \ cos 2x$
  3. $\ tan2x$

  4. None of these.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Given $E\tan (x-30^{\circ})=J\tan (x+120^{\circ})$

$\implies \dfrac{E}{J}=\dfrac{\tan (x+120^{\circ})}{\tan (x-30^{\circ})}$

Applying compoundo and dividendo rule

$\dfrac{E+J}{E-J}=\dfrac{\tan (x-30^{\circ})+\tan (x+120^{\circ})}{\tan (x+120^{\circ})-\tan (x-30^{\circ})}=\dfrac{\frac{\sin (x-30^{\circ})}{\cos (x-30^{\circ})}+\frac{\sin (x+120^{\circ})}{\cos (x+120^{\circ})}}{\frac{\sin (x+120^{\circ})}{\cos (x+120^{\circ})}-\frac{\sin (x-30^{\circ})}{\cos (x-30^{\circ})}}$

                                                        $=\dfrac{\sin (x-30^{\circ})\cos(x+120^{\circ})+\sin (x+120^{\circ})\cos(x-30^{\circ})}{\sin (x+120^{\circ})\cos(x-30^{\circ})-\sin (x-30^{\circ})\cos (x+120^{\circ})}$

                                                       $=\dfrac{\sin (x-30^{\circ}+x+120^{\circ})}{\sin (x+120^{\circ}-x+30^{\circ})}$

                                                      $=\dfrac{\sin (90^{\circ}+2 x)}{\sin 150^{\circ}}$

                                                      $=2\cos 2 x$
Multiple choice using trigonometric tables trigonometric ratios of some specific angles trigonometric identities trigonometry maths

The value of $\tan 7\dfrac{1}{2}^{o}$ is equal to

  1. $\sqrt{6}+\sqrt{3}+\sqrt{2}-2$
  2. $\sqrt{6}-\sqrt{3}+\sqrt{2}-2$
  3. $\sqrt{6}-\sqrt{3}+\sqrt{2}+2$
  4. $\sqrt{6}-\sqrt{3}-\sqrt{2}-2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using the formula tan(x/2) = sqrt((1-cos x)/(1+cos x)) with x = 15 degrees, or using the half-angle identity for 7.5 degrees, the value is sqrt(6) + sqrt(3) + sqrt(2) - 2.