Math, asked by agadre412, 13 hours ago

if sin thetha =x ^2 - y^2 /x^2+ y^2 then tan theta = ?​

Answers

Answered by user0888
5

\rm\large\underline{\text{Topic: Trigonometric identity}}

We know that,

\cdots\longrightarrow \sin^{2}\theta+\cos^{2}\theta=1.

After dividing both sides by \sin^{2}\theta,

\cdots\longrightarrow 1+\tan^{2}\theta=\dfrac{1}{\sin^{2}\theta}.

\rm\large\underline{\text{Explanation}}

Let's solve the question.

We know that,

\cdots\longrightarrow 1+\tan^{2}\theta=\dfrac{1}{\sin^{2}\theta}.

It is given that,

\cdots\longrightarrow\sin\theta=\dfrac{x^2-y^2}{x^2+y^2}

So,

\cdots\longrightarrow\dfrac{1}{\sin^{2}\theta}=\left(\dfrac{x^2+y^2}{x^2-y^2}\right)^{2}

And,

\cdots\longrightarrow1+\tan^{2}\theta=\left(\dfrac{x^2+y^2}{x^2-y^2}\right)^{2}

\cdots\longrightarrow \tan^{2}\theta=\left(\dfrac{x^2+y^2}{x^2-y^2}\right)^{2}-1

\cdots\longrightarrow \tan^{2}\theta=\left(\dfrac{x^2+y^2}{x^2-y^2}\right)^{2}-\left(\dfrac{x^{2}-y^{2}}{x^{2}-y^{2}}\right)^{2}

\cdots\longrightarrow \tan^{2}\theta=\left(\dfrac{2x^{2}+2y^{2}}{x^{2}-y^{2}}\right)^{2}

Hence,

\cdots\longrightarrow \boxed{\tan\theta=\dfrac{2x^{2}+2y^{2}}{x^{2}-y^{2}}}

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