Mathematics

Trigonometric Identities

87 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 trigonometric equations trigonometric functions trigonometry maths

General solution of $\dfrac{1-{tan}^{2}x}{{sec}^{2}x}=\dfrac{1}{2}$ is

  1. $n\pi+\dfrac{\pi}{6},n\in Z$
  2. $n\pi-\dfrac{\pi}{6},n\in Z$
  3. $n\pi\pm\dfrac{\pi}{6},n\in Z$
  4. $2n\pi\pm\dfrac{\pi}{6},n\in Z$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given $\dfrac{1-\tan^2 x}{\text{sec}^2 x}=\dfrac{1}{2}$


$\implies \dfrac{1-\tan^2 x}{1+\tan^2 x}=\dfrac{1}{2}$


$\implies 2-2\tan^2 x=1+\tan^2 x$

$\implies \tan^2 x=\dfrac{1}{3}=\tan^2 \dfrac{\pi}{6}$

$\implies x=n\pi\pm \dfrac{\pi}{6}$

Multiple choice trigonometric equations trigonometric functions trigonometry maths

If $3sin\alpha =5sin\beta ,\quad then\quad \frac { \tan { \frac { \alpha +\beta }{ 2 } } }{ \tan { \frac { \alpha -\beta }{ 2 } } } $ is equal to

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

Given $3\sin \alpha=5\sin \beta\implies \dfrac{\sin \alpha}{\sin \beta}=\dfrac{5}{3}$


Applying componendo and dividendo rule


$\implies \dfrac{\sin \alpha+\sin \beta}{\sin \alpha-\sin \beta}=\dfrac{5+3}{5-2}$

$\implies \dfrac{2\sin \dfrac{\alpha+\beta}{2}\cos \dfrac{\alpha-\beta}{2}}{2\sin \dfrac{\alpha-\beta}{2}\cos \dfrac{\alpha+\beta}{2}}=\dfrac{8}{2}$

$\implies \dfrac{\tan \dfrac{\alpha+\beta}{2}}{\tan \dfrac{\alpha-\beta}{2}}=4$

Multiple choice trigonometric equations trigonometric functions trigonometry maths

If $y\tan (A+B+C)=x\tan (A+B-C)=\lambda$, then $\tan 2C=?$

  1. $\dfrac{\lambda(x+y)}{\lambda^2-xy}$
  2. $\dfrac{\lambda(x+y)}{\lambda^2+xy}$
  3. $\dfrac{\lambda(x-y)}{xy-\lambda^2}$
  4. $\dfrac{\lambda (x-y)}{xy+\lambda^2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given $y\tan (A+B+C)=x\tan (A+B-C)=\lambda$

$\implies \tan (A+B+C)=\dfrac{\lambda}{y},\tan (A+B-C)=\dfrac{\lambda}{x}$

$\tan 2 C=\tan ((A+B+C)-(A+B-C))=\dfrac{\tan (A+B+C)-\tan (A+B-C)}{1+\tan (A+B+C)\tan (A+B-C)}$

                                                                                $=\dfrac{\frac{\lambda}{y}-\frac{\lambda}{x}}{1+\frac{\lambda^2}{x y}}$

                                                                                $=\dfrac{\lambda(x-y)}{x y+\lambda^2}$

Multiple choice trigonometric equations trigonometric functions trigonometry maths

If $4^{2\, sin^2x}.16^{tan^2x}.2^{4\, cos^2x} = 256 $ such that $0 < x < \dfrac{\pi}{2}$ then $x$ is equal to ___________.

  1. $\dfrac{\pi}{3}$
  2. $\dfrac{\pi}{4}$
  3. $\dfrac{\pi}{12}$
  4. $\dfrac{\pi}{24}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
${4}^{2{\sin}^{2}{x}}.{16}^{{\tan}^{2}{x}}.{2}^{4{\cos}^{2}{x}}=256$

$\Rightarrow\,{2}^{4{\sin}^{2}{x}}.{2}^{4{\tan}^{2}{x}}.{2}^{4{\cos}^{2}{x}}={2}^{8}$

$\Rightarrow\,{2}^{4{\sin}^{2}{x}+4{\tan}^{2}{x}+4{\cos}^{2}{x}}={2}^{8}$

$\Rightarrow\,4{\sin}^{2}{x}+4{\tan}^{2}{x}+4{\cos}^{2}{x}=8$

$\Rightarrow\,{\sin}^{2}{x}+{\tan}^{2}{x}+{\cos}^{2}{x}=2$

$\Rightarrow\,\left({\sin}^{2}{x}+{\cos}^{2}{x}\right)+{\tan}^{2}{x}=2$

$\Rightarrow\,1+{\tan}^{2}{x}=2$ since $\left({\sin}^{2}{x}+{\cos}^{2}{x}=1\right)$

$\Rightarrow\,{\sec}^{2}{x}=2$ since $1+{\tan}^{2}{x}={\sec}^{2}{x}$

$\Rightarrow\,{\cos}^{2}{x}=\dfrac{1}{2}$

$\Rightarrow\,\cos{x}=\pm\dfrac{1}{\sqrt{2}}$

$\Rightarrow\,\cos{x}=\dfrac{1}{\sqrt{2}}$ since $0<x<\dfrac{\pi}{2}$

$\Rightarrow\,x=\dfrac{\pi}{4}$

Multiple choice mathematics and statistics angle and their measurement degree measure of angle measure of angle radians or degrees

The value of $\tan\left(7\dfrac{1}{2}\right)^o$ is 

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

$\tan\left(7\dfrac{1}{2}\right)^o=\tan\left(\dfrac{15}{2}\right)^o$


                       $=\dfrac{\sin\dfrac{15^o}{2}}{\cos\dfrac{15^o}{2}}$

                       $=\dfrac{2\sin\dfrac{15^o}{2}\times \sin\dfrac{15^o}{2}}{2\sin\dfrac{15^o}{2}\times \cos\dfrac{15^o}{2}}$

                        $=\dfrac{2\sin^2\dfrac{15^o}{2}}{\sin\left(2\times\dfrac{15^o}{2}\right)}$

                        $=\dfrac{1-\cos\left(2\times\dfrac{15^o}{2}\right)}{\sin 15^o}$

                        $=\dfrac{1-\cos 15^o}{\sin 15^o}$

Now,
$\cos 15^o=\cos(45^o-30^o)$
              $=\cos 45^o\cos30^o+\sin 45^o\sin 30^o$
              $=\dfrac{1}{\sqrt{2}}\times\dfrac{\sqrt{3}}{2}+\dfrac{1}{\sqrt{2}}\times\dfrac {1}{2}$
              $=\dfrac{\sqrt{3}+1}{2\sqrt{2}}$

$\sin 15^o=\sin(45^o-30^o)$
             $=\sin 45^o\cos30^o-\cos 45^o\sin 30^o$
             $=\dfrac{1}{\sqrt{2}}\times\dfrac{\sqrt{3}}{2}-\dfrac{1}{\sqrt{2}}\times\dfrac {1}{2}$
             $=\dfrac{\sqrt{3}-1}{2\sqrt{2}}$


$\tan\left(7\dfrac{1}{2}\right)^o=\dfrac{1-\dfrac{\sqrt{3}+1}{2\sqrt{2}}}{\dfrac{\sqrt{3}-1}{2\sqrt{2}}}$

                      $=\dfrac{2\sqrt{2}-(\sqrt{3}+1)}{\sqrt{3}-1}$

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 }  } $
$x=\left( 2n+1 \right) \cfrac { \pi  }{ 2 } ,n\in I$