Tag: trigonometric functions

Quizzes (5) Questions (20)

Questions Related to trigonometric functions

Multiple choice trigonometric equations trigonometric functions trigonometry maths

If $\tan 4x+\tan 5x-\tan 9x=k\tan 4x\tan 5x\tan 9x$ then $k=$

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

$\begin{array}{l}\tan 9x = \tan (4x + 5x) = \dfrac{{\tan 4x + \tan 5x}}{{1 - \tan 4x\tan 5x}}\ \Rightarrow \tan 9x - \tan 4x\tan 5x\tan 9x = \tan 4x + \tan 5x\ \Rightarrow \tan 4x + \tan 5x - \tan 9x =  - \tan 4x\tan 5x\tan 9x\\therefore k =  - 1\end{array}$

Multiple choice trigonometric equations trigonometric functions trigonometry maths

State true or false $\tan(\dfrac{\pi}{4} + \theta) - \tan(\dfrac{\pi}{4} -\theta) = 2\tan\theta$

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Using $\tan (A+B)=\cfrac {\tan A+\tan B}{1-\tan A \tan B}$

$\Rightarrow \tan (A-B)=\cfrac {\tan A-\tan B}{1+\tan A+\tan B}$
$\tan \left( \cfrac {\pi}{4}+\theta\right)-\tan \left(\cfrac {\pi}{4}-\theta\right)$
$\Rightarrow A=\cfrac {\pi}{4}, B=\theta$
$\Rightarrow \cfrac {\tan \cfrac {\pi}{4}+\tan\theta}{1-\tan \cfrac {\pi}{4}\tan \theta}-\cfrac {\tan \cfrac {\pi}{4}-\tan \theta}{1+\tan \cfrac {\pi}{4}\tan \theta}$
$\because \tan \cfrac {\pi}{4}=1$
$\Rightarrow \cfrac {1+\tan\theta}{1-\tan \theta}-\cfrac {1-\tan \theta}{1+\tan \theta}$
$\Rightarrow \cfrac {(1+\tan\theta)^2-(1-\tan\theta)^2}{(1-\tan\theta)(1+\tan\theta)}$
$\therefore a^2-b^2=(a+b)(a-b)$
$\Rightarrow \cfrac {1+\tan^2\theta+2\tan\theta-1-\tan^2\theta+2\tan\theta}{1-\tan^2\theta}$
$\Rightarrow \cfrac {4\tan\theta}{1-\tan^2\theta}$
$\because \tan2\theta=\cfrac {2\tan\theta}{1-\tan^2\theta}$
$=2\tan2\theta$.

Multiple choice trigonometric equations trigonometric functions trigonometry maths

State true or false

$\tan{ 18 }^{ 0 }+\tan27^{ 0 }+\tan{ 18 }^{ 0 }\cdot \tan27^{ 0 }=1$

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$tan180+tan270+tan180+tan270=1$

$0.32491969 + 0.50952544+0.32491969 + 0.50952544 = 1$

$1.66889026 = 1$

The left side $1.66889026 $ not equal to rightside$ 1$.

Which means the given statement is false.

 

Multiple choice trigonometric equations trigonometric functions trigonometry maths

$tan  5x-tan  3x-tan  2x=$

  1. $\tan 5x \tan 3x \tan 2x$
  2. $\sin 5x \sin 3x \sin 2x$
  3. $\cos 5x \cos 3x \cos 2x$
  4. $\sec 5x \sec 3x \sec 2x$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

We've,

$\tan (3x+2x)=\tan 5x$

or, $\dfrac{\tan 3x+\tan 2x}{1-\tan 3x.\tan 2x}=\tan 5x$

or, $\tan 3x+\tan 2x=\tan 5x-\tan 2x.\tan 3x.\tan 5x$

or, $\tan 5x-\tan 3x-\tan 2x=\tan 2x.\tan 3x.\tan 5x$.

Multiple choice trigonometric equations trigonometric functions trigonometry maths

$A, B, C$ are three angles such that $\tan  A+\tan  B+\tan  C=\tan  A  \tan  B  \tan  C.$ Which of the following statements is always correct ?

  1. $ABC$ is a triangle, i.e. $A+B+C=\pi $
  2. $A=B=C. i.e., $ $ABC$ is an equilateral triangle
  3. $A+B=C, $ i.e., $ABC$ is a right- angled triangle
  4. $A+B=\pi $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

(A) $tan\left [ (A+B)+C \right ]$
$=\frac{tan (A+B)+tan C}{1-tan (A+B)  tan  C}=\frac{\frac{tan  A+tan   B}{1-tan  A   tan  B}+tan  C}{1-\frac{tan  A+tan  B}{1-tan   A   tan  B}.  tan  C}$
$=\frac{tan  A+tan  B+tan  C-tan  A   tan  B   tan  C}{Denominator}$
$=0$
$\left [ since,  tan  A+tan  B+tan  C =tan  A   tan  B   tan  C \right ]$
$\therefore A+B+C=\pi $ i.e.,  A, B, C is a triangle

Multiple choice trigonometric equations trigonometric functions trigonometry maths

If $\dfrac{\pi}{4}<A<\dfrac{\pi}{2}$ then $\tan^{-1}\left(\dfrac{1}{2}\tan 2A\right)+\tan^{-1}(\cot A)+\tan^{-1}(\cot^{3}A)$=

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

Using trigonometric identities for inverse functions and the given range, the sum simplifies to 0. Specifically, the terms cancel out based on the properties of inverse tangents of cotangent functions.

Multiple choice trigonometric equations trigonometric functions trigonometry maths

If $A+B+C=\pi $ and cosA=cosB cosC, then tanB tanC is equal to 

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

Given $A+B+C=\pi\implies A=\pi-(B+C)$

And also given $\cos A=\cos B\cos C$
$\implies \cos (\pi-(B+C))=\cos B\cos C$
$\implies -\cos B\cos C+\sin B\sin C=\cos B\cos C$
$\implies \sin B\sin C=2\cos B\cos C$
$\implies \tan B\tan C=2$

Multiple choice trigonometric equations trigonometric functions trigonometry maths

$\alpha, \beta$ are the solution (s) of $3 cos 2 \theta + 4 sin 2 \theta = 5$
$tan (\alpha + \beta) = $

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

The equation 3cos(2theta) + 4sin(2theta) = 5 can be written as 5(3/5 cos(2theta) + 4/5 sin(2theta)) = 5, or cos(2theta - phi) = 1, where tan(phi) = 4/3. The solutions alpha and beta lead to tan(alpha + beta) = 1.

Multiple choice trigonometric equations trigonometric functions trigonometry maths

$\alpha, \beta$ are the solution (s) of $3 cos 2 \theta + 4 sin 2 \theta = 5$
$tan (\alpha - \beta) = $

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

Since 3cos(2theta) + 4sin(2theta) = 5 has only one solution for 2theta in the range [0, 2pi), alpha and beta are essentially the same value (or differ by a multiple of pi), making tan(alpha - beta) = 0.