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

Multiple choice using trigonometric tables trigonometric ratios of some specific angles trigonometric identities trigonometry maths

The value of expression $\dfrac { 2\left( \sin{ 1 }^{ o }+\sin{ 2 }^{ o }+\sin{ 3 }^{ o }+.....+\sin{ 89 }^{ o } \right)  }{ 2\left( \cos{ 1 }^{ o }+\cos{ 2 }^{ o}+......+\cos{ 44 }^{ o } \right) +1 }$ equals

  1. $\sqrt{2}$
  2. $1/\sqrt{2}$
  3. $1/2$
  4. $0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The numerator is a sum of sines from 1 to 89, which equals cot(1/2) * sin^2(45). The denominator simplifies similarly. The ratio evaluates to sqrt(2).

Multiple choice direction cosines and direction ratios three dimensional geometry maths

If the points $a(cos \alpha + i sin \alpha)$ , $b(cos \beta + i sin \beta)$ and $c(cos \gamma + isin \gamma)$ are collinear then the value of $|z|$ is:  
( where ${z = bc  \ sin(\beta-\gamma) + ca \ sin(\gamma-\alpha) + ab \ sin(\alpha - \beta) + 3i -4k}$ )

  1. $2$
  2. $5$
  3. $1$

  4. None of these.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Given $a=cos\alpha+i\sin\alpha=e^{i\alpha}$ , $b=cos\beta+isin\beta=e^{i\beta}$ and $c=cos\gamma+isin\gamma=e^{i\gamma}$
Consider $bcsin(\beta-\gamma)=e^{i(\beta+\gamma)}sin(\beta-\gamma)=\frac{1}{2i}e^{i(\beta+\gamma)}(e^{i(\beta-\gamma)}-e^{-i(\beta-\gamma)}) = \frac{1}{2i}(e^{i(2\beta)}-e^{i(2\gamma)})$
Similarly we get $casin(\gamma-\alpha) = \frac{1}{2i}(e^{i(2\gamma)}-e^{i(2\alpha)})$ and $absin(\alpha-\beta) = \frac{1}{2i}(e^{i(2\alpha)}-e^{i(2\beta)})$
Therefore we get $bcsin(\beta-\gamma)+casin(\gamma-\alpha)+absin(\alpha-\beta)=0$
So we get $z=3i-4i$
$\Rightarrow |z|=5$
Multiple choice maths squares and square roots finding the square of a number finding square of a number patterns in square numbers

If sin$\theta -cosec  \theta =\sqrt{5},$ then the value of sin  $\theta  + cosec  \theta$ is:

  1. $\sqrt{3}$

  2. 1

  3. 3

  4. 9
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\Rightarrow \sin\theta-cosec\theta=\sqrt{5}$


$\Rightarrow \sin\theta-\dfrac{1}{\sin\theta}=\sqrt{5}$      $(\because cosec\theta=\dfrac{1}{\sin\theta})$

$\Rightarrow \sin^2\theta-\sqrt{5}\sin\theta-1=0$

Solving equation to get roots.

$\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}=\dfrac{\pm3+\sqrt{5}}{2}$ (substitute values to get roots)

To find:-

$\sin\theta+cosec\theta$

Using $\dfrac{3+\sqrt{5}}{2}$

$=\dfrac{3+\sqrt{5}}{2}+\dfrac{2}{3+\sqrt{5}}$

$=\dfrac{(3+\sqrt{5})^2+4}{2(3+\sqrt{5})}$

$=\dfrac{9+4+5+6\sqrt{5}}{2(3+\sqrt{5})}$

$=\dfrac{6(3+\sqrt{5})}{2(3+\sqrt{5})}$

$=3$


Using $\dfrac{-3+\sqrt{5}}{2}$

$=\dfrac{-3+\sqrt{5}}{2}+\dfrac{2}{-3+\sqrt{5}}$

$=\dfrac{(-3+\sqrt{5})^2+4}{2(-3+\sqrt{5})}$

$=\dfrac{9+4+5-6\sqrt{5}}{2(-3+\sqrt{5})}$

$=\dfrac{-6(-3+\sqrt{5})}{2(-3+\sqrt{5})}$

$=-3$


According to option answer is $3$

Multiple choice business maths applications of matrices and determinants non-homogeneous linear equations system of simultaneous equations matrices

The values of $\theta $ lying between $\theta =0$ and $\theta =\dfrac {\pi}{2}$ and satisfying the equation
$\begin{vmatrix}
1+\sin ^{2}\theta  & \cos ^{2}\theta  & 4\sin 6\theta \
\sin ^{2}\theta  & 1+\cos ^{2}\theta  & 4\sin 6\theta \
\sin ^{2}\theta  & \cos ^{2}\theta  & 1+4\sin 6\theta
\end{vmatrix}$
are given by

  1. $\dfrac {\pi }{36}, \dfrac{5\pi}{ 36}$
  2. $\dfrac{7\pi}{36}, \dfrac{11\pi}{3}$
  3. $\dfrac{5\pi }{36}, \dfrac{7\pi }{36}$
  4. $\dfrac{11\pi}{36}, \dfrac{\pi }{36}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\begin{vmatrix} 1+\sin ^{ 2 } \theta  & \cos ^{ 2 } \theta  & 4\sin  6\theta  \ \sin ^{ 2 } \theta  & 1+\cos ^{ 2 } \theta  & 4\sin  6\theta  \ \sin ^{ 2 } \theta  & \cos ^{ 2 } \theta  & 1+4\sin  6\theta  \end{vmatrix}=0$

Applying ${ R } _{ 3 }\rightarrow { R } _{ 3 }-{ R } _{ 1 },{ R } _{ 2 }\rightarrow { R } _{ 2 }-{ R } _{ 1 }$

$\Rightarrow \begin{vmatrix} 1+\sin ^{ 2 } \theta  & \cos ^{ 2 } \theta  & 4\sin  6\theta  \ -1 & 1 & 0 \ -1 & 0 & 1 \end{vmatrix}=0$

Applying ${ C } _{ 1 }\rightarrow { C } _{ 1 }+{ C } _{ 2 }$

$\Rightarrow \begin{vmatrix} 2 & \cos ^{ 2 } \theta  & 4\sin  6\theta  \ 0 & 1 & 0 \ -1 & 0 & 1 \end{vmatrix}=0\ \Rightarrow 2+4\sin  6\theta =0\Rightarrow \sin  6\theta =-\cfrac { 1 }{ 2 } \ \Rightarrow 6\theta =n\pi +{ \left( -1 \right)  }^{ n }\left( -\cfrac { \pi  }{ 6 }  \right) \ \Rightarrow \theta =\cfrac { n\pi  }{ 6 } +{ \left( -1 \right)  }^{ n+1 }\left( \cfrac { \pi  }{ 36 }  \right) \ \Rightarrow \theta =\cfrac { 7\pi  }{ 36 } ,\cfrac { 11\pi  }{ 36 } $

Multiple choice business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

If $A = \left[ \begin{array}{l}\cos \theta \,\,\,\,\sin \theta \ - \sin \theta \,\,\,\cos \theta \end{array} \right]$ where $\theta  = \frac{{2\pi }}{{19}}$ then ${A^{2017}} = $

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

A is a rotation matrix R(theta). A^n = R(n*theta). Here, A^2017 = R(2017 * 2pi / 19). Since 2017 = 19 * 106 + 3, A^2017 = R(3 * 2pi / 19) = A^3.