Tag: trigonometrical ratio and identities

Quizzes (3) Questions (2)

Questions Related to trigonometrical ratio and identities

Multiple choice maths trigonometric ratios of acute angles compound angles, multiple angles, sub multiple angles and transformation formulae trigonometric identities trigonometrical ratio and identities

$\sin ^{ 8 }{ \theta  } -\cos ^{ 8 }{ \theta  } -\left( \sin ^{ 2 }{ \theta  } -\cos ^{ 2 }{ \theta  }  \right) \left( 1-\sin ^{ 2 }{ \theta  }  \right) $=0

  1. True

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

$\sin^4 \theta+\cos^4 \theta=(\sin^2 \theta+\cos^2 \theta)^2-2\sin^2 \theta\cos^2 \theta=1-2\sin^2 \theta\cos^2 \theta$

$\sin^4 \theta-\cos^4 \theta=(\sin^2 \theta-\cos^2 \theta)(\sin^2 \theta+\cos^2 \theta)=\sin^2 \theta-\cos^2 \theta$
$\sin^8\theta-\cos^8\theta-(\sin^2 \theta-\cos^2 \theta)(1-\sin^2 \theta)=(\sin^4 \theta+\cos^4 \theta)(\sin^4 \theta-\cos^4 \theta)-(\sin^2 \theta-\cos^2 \theta)(1-\sin^2 \theta)$
                                                                                 $=(1-2\sin^2 \theta\cos^2 \theta)(\sin^2 \theta-\cos^2 \theta)-(\sin ^2 \theta-\cos^2\theta)(1-\sin^2 \theta)$
                                                                                 $=(\sin^2 \theta-\cos^2 \theta)(\sin^2 \theta)(1-2\cos^2 \theta)$
So the given relation is $\text{False}$

Multiple choice maths compound angles, multiple angles, sub multiple angles and transformation formulae trigonometric ratios of acute angles trigonometric identities trigonometrical ratio and identities

If $tan x + cot x = 2$, then $sin^{2n}x+cos^{2n}x=$

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

Given $\tan x+\cot x=2$

$\implies \tan x+\dfrac{1}{\tan x}=2$
$\implies \tan^2 x-2\tan x+1=0$
$\implies (\tan x-1)^2=0$
$\implies \tan x=1\implies x=\dfrac{\pi}{4}$
$\sin^{2 n} x+\cos^{2 n} x=\bigg(\dfrac{1}{\sqrt{2}}\bigg)^{2 n}+\bigg(\dfrac{1}{\sqrt{2}}\bigg)^{2 n}=\dfrac{1}{2^n}+\dfrac{1}{2^n}=\dfrac{2}{2^n}=\dfrac{1}{2^{n-1}}$