Mathematics

Trigonometric Identities and Equations

223 Questions

Solve a variety of questions based on trigonometric identities and mathematical equations. Key areas covered include double angle formulas, the law of sines, and quadrant angles. This material helps students preparing for advanced mathematics exams, UPSC, and state public service commissions.

Double angle formulasLaw of sinesTrigonometric quadrantsLaplace transformSecant functionTaylor series

Trigonometric Identities and Equations Questions

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

If $11 \sin^2 x + 7\cos^2x = 8$ then $x =$______

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

Given $11\sin^2 x+7\cos^2 x=8$

$\implies 11\sin^2 x+7-7\sin^2 x=8$
$\implies 4\sin^2 x=1$
$\implies \sin^2 x=\dfrac{1}{4}$
$\implies \sin^2 x=\sin^2 \dfrac{\pi}{6}$
$\implies x=n\pi\pm \dfrac{\pi}{6},\forall Z$

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

If $\alpha, \beta$ are solution of equation a $cos \theta + b sin\theta = c$ then

  1. $sin \alpha + sin \beta = \dfrac{a^2-c^2}{b^2-a^2}$
  2. $cos \alpha + cos \beta = \dfrac{2ac}{a^2 + b^2}$
  3. $cos \alpha . cos \beta = \dfrac{c^2-b^2}{a^2 + b^2}$
  4. $\sin \alpha.\sin \beta=\dfrac{a^{2}-c^{2}}{b^{2}-a^{2}}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
$a\cos\theta+b\sin\theta=c$

$a^{2}\cos^{2}\theta+b^{2}\sin^{2}\theta+2ab\sin^{2}\theta.\cos\theta=c^{2}$

$a^{2}(1-\sin^{2}\theta)+b^{2}\sin^{2}\theta+2ab\sin\theta.\cos\theta=c^{2}$

$a^{2}-a^{2}\sin^{2}\theta+b^{2}\sin^{2}\theta\sin^{2}\theta+2ab\sin\theta\cos\theta=c^{2}$

$\sin^{2}\theta(b^{2}-a^{2})+2ab\sin\theta.\cos\theta+a^{2}-c^{2}=0$

So
$\sin \alpha.\sin \beta=\dfrac{a^{2}-c^{2}}{b^{2}-a^{2}}$
Multiple choice mathematics and statistics angle and their measurement angles and sides naming the sides in a right angled triangle understanding ratios

If $\cos x + cosy + \cos \theta = 0$ and $\sin x + \sin y + \sin \theta = 0$, then $\cot\left(\dfrac{x + y}{2}\right)$ 

  1. $\sin \theta$
  2. $\cos \theta$
  3. $\cot \theta$
  4. $\sin\left(\dfrac{x + y}{2}\right)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\cos x+\cos y+\cos \theta=0$

$\cos x+\cos y=-\cos \theta$
$\cfrac { 2\cos(x+y) }{ 2 } \times \cfrac { \cos(x-y) }{ 2 } =-\cos\theta  \longrightarrow 1$
$\sin x+\sin y+\sin \theta=0$
$\sin x+\sin y=-\sin \theta$
$\cfrac { 2\sin(x+y) }{ 2 } \times \cfrac { \sin(x-y) }{ 2 } =-\sin\theta $
Dividing both,
$\cfrac { \cfrac { 2\cos  (x+y) }{ 2 } \times \cfrac { \cos  (x-y) }{ 2 }  }{ \cfrac { 2\sin  (x+y) }{ 2 } \times \cfrac { \sin  (x-y) }{ 2 }  } =\cfrac { -\cos  \theta  }{ -\sin\theta } $
$\cfrac { \cot(x+y) }{ 2 } =\cot\theta $

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

If $sin:\theta +cos:\theta =p$ and $:tan:\theta +cot:\theta =q$ then $:q\left(p^2-1\right)=$

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

$sin:\theta +cos:\theta =p$ 
Squaring on both side we get
$\Rightarrow 1+sin:2\theta =p^2$    $[\because sin:2\theta =p^2-1]$
$Tan\theta +\frac{1}{Tan\theta :}=q$

$\frac{Tan^2\theta +1}{2Tan\theta :}=\frac{q}{2}$

$cosec:2\theta =\frac{q}{2}$

$\Rightarrow \frac{1}{p^2-1}=\frac{q}{2}$   $\Rightarrow 2=\left(p^2-1\right)q$

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

If $\tan { \theta  } .\tan { (120-\theta ) } .\tan { (120+\theta ) } =\dfrac { 1 }{ \sqrt { 3 }  }$, then $\theta $

  1. $\dfrac { n\pi }{ 3 } +\dfrac { \pi }{ 18 } ,n\epsilon Z$
  2. $\dfrac { n\pi }{ 3 } +\cfrac { \pi }{ 12 } ,n\epsilon Z$
  3. $\dfrac { n\pi }{ 12 } +\dfrac { \pi }{ 12 } ,n\epsilon Z$
  4. $\dfrac { n\pi }{ 3 } +\dfrac { \pi }{ 6 } ,n\epsilon Z$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using the identity tan(theta) * tan(60-theta) * tan(60+theta) = tan(3*theta), we have tan(3*theta) = 1/sqrt(3). Thus, 3*theta = n*pi + pi/6, which simplifies to theta = n*pi/3 + pi/18.

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

In $\Delta ABC$, a, b, c are the lengths of its sides and A, B, C are the angles of triangle ABC. The correct relation is 

  1. $(b-c)sin(\frac{B-C}{2}) =a cos(\frac{A}{2}) $
  2. $(b-c)cos(\frac{A}{2})= a sin(\frac{B-C}{2}) $
  3. $(b+c)sin(\frac{B+C}{2})=a cos(\frac{A}{2}) $
  4. $(b-c)cos(\frac{A}{2}) = 2a sin(\frac{B+C}{2}) $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Using the Law of Sines and Mollweide's formulas, the correct relation is (b-c)cos(A/2) = a*sin((B-C)/2).

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

The value of sin $15^0$ is

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

$\sin15^o$


$=\sin(45^o-30^o)$

$=\sin45^o \ \cos30^o - \cos45^o \ \sin30^o$

$=\dfrac{1}{\sqrt{2}} \cdot \dfrac{\sqrt{3}}{2}-\dfrac{1}{\sqrt{2}} \cdot \dfrac{1}{2}$

$=\dfrac{\sqrt{3}-1}{2\sqrt{2}}$
Hence answer is D

Multiple choice the nth roots of unity complex numbers maths

The value of $\sum _{ n=1 }^{ 10 }{ \left( sin\frac { 2n\pi  }{ 11 } -icos\frac { 2n\pi  }{ 11 }  \right)  } $

  1. $i$
  2. $-i$
  3. $0$

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

The sum is a geometric series of complex numbers. Using the identity for the sum of roots of unity, the sum evaluates to i.

Multiple choice mathematics and statistics introduction to set union and intersections union and intersection of sets basic operations on sets

Let $P={ \theta :sin\theta -cos\theta =\sqrt { 2 } cos\theta } $ and $Q={ sin\theta + cos\theta =\sqrt { 2 } sin\theta } $ be two sets. Then:

  1. $P\subset Q\quad and\quad Q-P\neq \emptyset $
  2. $Q\subset P$
  3. $P\subset Q$
  4. $P=Q$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$P = \left{ {\theta :\sin \theta  - \cos \theta  = \sqrt 2 \cos \theta } \right}$

$Q = \left{ {\theta :\sin \theta  + \cos \theta  = \sqrt 2 \sin \theta } \right}$
From $P$
$\sin \theta  - \cos \theta  = \sqrt 2 \cos \theta $
$\sin \theta  = \left( {\sqrt 2  + 1} \right)\cos \theta $
$\frac{{\sin \theta }}{{\cos \theta }} = \left( {\sqrt 2  + 1} \right)$
$\tan \theta  = \left( {\sqrt 2  + 1} \right)$
from $Q$
$\sin \theta  + \cos \theta  = \sqrt 2 \sin \theta $
$\sin \theta \left( {\sqrt 2  - 1} \right) = \cos \theta $
$\frac{{\sin \theta }}{{\cos \theta }} = \left( {\sqrt 2  - 1} \right)$
$\tan \theta  = \left( {\sqrt 2  - 1} \right)$
$\tan \theta  = \frac{1}{{\sqrt 2  - 1}} \times \frac{{\sqrt 2  + 1}}{{\sqrt 2  + 1}} = \sqrt 2  + 1$
$\therefore P = Q$
Hence,
option $(D)$ is correct answer.

Multiple choice maths surface area and volume of cube and cuboid length of the diagonal diagonal of cube and cuboid surface area of cubes and cuboids

Evaluate cos$\begin{pmatrix}2csc^{-1}(\dfrac{x+4}{5})\end{pmatrix} = $

  1. $\dfrac{x^2+8x-16}{x+4}$
  2. $\dfrac{x^2+8x-16}{(x+4)^2}$
  3. $\dfrac{x^2+8x-34}{x+4}$
  4. $\dfrac{x^2+8x-34}{(x+4)^2}$
  5. $\dfrac{-16-8x-x^2}{(x+4)^2}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let $\theta =\csc ^{ -1 }{ \left( \dfrac { x+4 }{ 5 }  \right)  }$ implies that $\csc { (\theta )=\dfrac { x+4 }{ 5 }  }$ and therefore, $\sin { (\theta ) } =\dfrac { 5 }{ x+4 }$.


Use the pythagorean theorem to determine that the remaining leg of the right triangle has length:

$\sqrt { (x+4)^{ 2 }-5^{ 2 } } =\sqrt { x^{ 2 }+16+8x-25 } =\sqrt { x^{ 2 }+8x-9 }$

Therefore, $\cos { (\theta ) } =\dfrac { \sqrt { x^{ 2 }+8x-9 }  }{ x+4 }$  

Hence, $\cos { (2\theta ) } =\cos ^{ 2 }{ (\theta ) } -\sin ^{ 2 }{ (\theta ) } $
$=\left( \dfrac { \sqrt { x^{ 2 }+8x-9 }  }{ x+4 }  \right) ^{ 2 }-\left( \dfrac { 5 }{ x+4 }  \right) ^{ 2 }$
$=\dfrac { x^{ 2 }+8x-9 }{ \left( x+4 \right) ^{ 2 } } -\dfrac { 25 }{ \left( x+4 \right) ^{ 2 } } $
$=\dfrac { x^{ 2 }+8x-34 }{ \left( x+4 \right) ^{ 2 } }$ 

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

if $\displaystyle Sin\theta =\frac{3}{5}$ what is the value of $\displaystyle  \left ( \tan \theta +\sec \theta  \right )^{2}$?

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

$Sin \theta=\dfrac{P}{H}=\dfrac{3}{5}$

According to the Pythagorean therom
$H^2=B^2+P^2$
$\Rightarrow B=\sqrt{h^2-p^2}$
$\Rightarrow B=\sqrt{5^2-3^2}$
$\Rightarrow B=\sqrt{16}$
$\Rightarrow B=4 cm$
$\therefore tan \theta= \dfrac{P}{B}=\dfrac{3}{4}$
$sec \theta=\dfrac{H}{B}=\dfrac{5}{4}$
$\therefore (tan \theta+sec \theta)^2=(\dfrac{3}{4}+\dfrac{5}{4})^2$
$\Rightarrow (\dfrac{8}{4})^2=(2)^2=4$

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

If $sin\theta = 3sin(\theta +2\alpha)$, then the value of $tan(\theta+\alpha)+ 2tan\alpha$ is

  1. 3

  2. 2

  3. 1

  4. 0
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
$\sin \theta =3\sin \left( \theta +2\alpha\right)$

$\Rightarrow 3\sin \left( \theta +\alpha +\alpha \right)=\sin \theta$

      $3\sin \left(\theta+\alpha\right)\cos \alpha +3\cos \left(\theta+\alpha\right)\sin \alpha =\sin \left(\theta+\alpha-\alpha \right)$

      $3\sin \left(\theta+\alpha\right)\cos\alpha +3\cos \left(\theta+\alpha\right)\sin \alpha  =\sin \left(\theta+\alpha\right) \cos \alpha -\sin \alpha \cos \left(\theta+\alpha\right)$

      $2\sin\left(\theta+\alpha\right)\cos \alpha =-4\cos \left(\theta+\alpha\right)\sin \alpha$

      $2\tan \left(\theta+\alpha\right)=-4\tan \alpha$

      $\tan \left(\theta+\alpha\right)=-2\tan \alpha$

$\Rightarrow \tan \left(\theta+\alpha\right)+2\tan \alpha=0$

Hence, the answer is $0.$

Multiple choice mathematics and statistics introduction to set union and intersections union and intersection of sets basic operations on sets

If $A={\theta :\tan \theta -\tan^2\theta > 0}, B={\theta :|\sin \theta | < 1/2}$ find $A\cap B$.

  1. $\left(0 , \cfrac{7\pi}{6}\right)$
  2. $\left(0 , \cfrac{\pi}{6}\right)$
  3. $\left(0 , -\cfrac{\pi}{6}\right)$

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

$\tan \theta - \tan ^2 \theta > 0$ is only true when 

$0<\tan \theta < 1$
or,$ 0< \theta < \cfrac{\pi}{4}$
$A = (0,\cfrac{\pi}{4})$
$|\sin \theta| < \cfrac{1}{2}$
$-\cfrac{1}{2} < \sin \theta < \cfrac{1}{2}$
$B = (-\cfrac{\pi}{6} , \cfrac{\pi}{6})$
$A \cap B = (0 , \cfrac{\pi}{6})$