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If given a trigonometric equation$\sqrt 3 \tan \theta = 3\sin \theta $, find the value of ${\sin ^2}\theta - {\cos ^2}\theta $

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Last updated date: 22nd Jul 2024
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Answer
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Hint: - Use the trigonometric identities and Pythagoras theorem.

Given:$\sqrt 3 \tan \theta = 3\sin \theta $
$
   \Rightarrow \tan \theta = \dfrac{3}{{\sqrt 3 }}\sin \theta \\
   \Rightarrow \tan \theta = \sqrt 3 \sin \theta \\
   \Rightarrow \dfrac{{\tan \theta }}{{\sin \theta }} = \sqrt 3 \\
   \Rightarrow \cos \theta = \dfrac{1}{{\sqrt 3 }} \\
   \Rightarrow \cos \theta = \dfrac{{Adjacent{\text{ }}side}}{{Hypotenuse}} = \dfrac{B}{H} = \dfrac{1}{{\sqrt 3 }} \\
 $
seo images

From the above figure for the Right angled triangle by using Pythagoras Theorem,
$
  {H^2} = {P^2} + {B^2} \\
  {\left( {\sqrt 3 } \right)^2} = {P^2} + {1^2} \\
  {P^2} = 3 - 1 \\
  {P^2} = 2 \\
  P = \sqrt 2 \\
$
Now, ${\sin ^2}\theta - {\cos ^2}\theta = {\left( {\dfrac{P}{H}} \right)^2} - {\left( {\dfrac{B}{H}} \right)^2}$
$
   = {\left( {\dfrac{{\sqrt 2 }}{{\sqrt 3 }}} \right)^2} - {\left( {\dfrac{1}{{\sqrt 3 }}} \right)^2} \\
   = \dfrac{2}{3} - \dfrac{1}{3} \\
   = \dfrac{1}{3} \\
$
Note: The above question can be solved by using trigonometric identities, but here it is done by visualizing the terms in the form of sides of the right angled triangle, thus making the problem easier to solve.