Question

find the value of theta if (sin² theta)÷ tan² theta - sin²theta = 3​

Answer 1

Step-by-step explanation:

We have,

$\frac{ \sin ^{2} ( \theta) }{ \tan ^{2} ( \theta) - \sin ^{2} ( \theta) } = 3$

$\implies\frac{ \sin ^{2} ( \theta) }{ \sin ^{2} ( \theta)( \frac{1}{ \cos ^{2} ( \theta) } - 1)} = 3$

$\implies \frac{ \cos^{2} ( \theta) }{ \sin^{2} ( \theta) } = 3$

$\implies \cot^{2} ( \theta) = 3$

$\implies \tan ^{2} ( \theta) = \frac{1}{3}$

$\implies \tan ^{2} ( \theta) = \tan ^{2} ( \frac{\pi}{6} )$

$\implies \theta = n\pi ±\frac{\pi}{6}$

Answer 2

Answer:

We have,

\begin{gathered} \frac{ \sin ^{2} ( \theta) }{ \tan ^{2} ( \theta) - \sin ^{2} ( \theta) } = 3 \\ \end{gathered}

tan

2

(θ)−sin

2

(θ)

sin

2

(θ)

=3

\begin{gathered} \implies\frac{ \sin ^{2} ( \theta) }{ \sin ^{2} ( \theta)( \frac{1}{ \cos ^{2} ( \theta) } - 1)} = 3 \\ \end{gathered}

⟹

sin

2

(θ)(

cos

2

(θ)

1

−1)

sin

2

(θ)

=3

\begin{gathered} \implies \frac{ \cos^{2} ( \theta) }{ \sin^{2} ( \theta) } = 3 \\ \end{gathered}

⟹

sin

2

(θ)

cos

2

(θ)

=3

\implies \cot^{2} ( \theta) = 3⟹cot

2

(θ)=3

\begin{gathered} \implies \tan ^{2} ( \theta) = \frac{1}{3} \\ \end{gathered}

⟹tan

2

(θ)=

3

1

\begin{gathered} \implies \tan ^{2} ( \theta) = \tan ^{2} ( \frac{\pi}{6} ) \\ \end{gathered}

⟹tan

2

(θ)=tan

2

(

6

π

)

\begin{gathered} \implies \theta = n\pi ±\frac{\pi}{6} \\\end{gathered}

⟹θ=nπ±

6

π