If given a trigonometric equation$\sqrt 3 \tan \theta = 3\sin \theta $, find the value of ${\sin ^2}\theta - {\cos ^2}\theta $
Answer
366.3k+ views
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 }} \\
$
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.
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 }} \\
$

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.
Last updated date: 01st Oct 2023
•
Total views: 366.3k
•
Views today: 8.66k
Recently Updated Pages
What do you mean by public facilities

Difference between hardware and software

Disadvantages of Advertising

10 Advantages and Disadvantages of Plastic

What do you mean by Endemic Species

What is the Botanical Name of Dog , Cat , Turmeric , Mushroom , Palm

Trending doubts
How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Difference Between Plant Cell and Animal Cell

Why are resources distributed unequally over the e class 7 social science CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Briefly mention the contribution of TH Morgan in g class 12 biology CBSE

What is the past tense of read class 10 english CBSE
