Question

if sin theta equal to a square minus b square by a square + b square find value of other trigonometric ratios

Answer 1

$\textbf{Given:}$

$\sin{\theta}=\dfrac{a^2-b^2}{a^2+b^2}$

$\textbf{To find:}$

$\text{The value of other trigonometric ratios}$

$\textbf{Solution:}$

$\text{Consider,}$

$\sin{\theta}=\dfrac{a^2-b^2}{a^2+b^2}$

$\text{Taking reciprocals, we get}$

$\boxed{\bf\csc{\theta}=\dfrac{a^2+b^2}{a^2-b^2}}$

$\text{Now}$

$\cos^2{\theta}=1-\sin^2{\theta}$

$\cos^2{\theta}=1-(\dfrac{a^2-b^2}{a^2+b^2})^2$

$\cos^2{\theta}=\dfrac{(a^2+b^2)^2-(a^2-b^2)^2}{(a^2+b^2)^2}$

$\cos^2{\theta}=\dfrac{(a^2+b^2+a^2-b^2)(a^2+b^2-a^2+b^2)}{(a^2+b^2)^2}$

$\cos^2{\theta}=\dfrac{(2a^2)(2b^2)}{(a^2+b^2)^2}$

$\implies\boxed{\bf\cos{\theta}=\dfrac{2ab}{a^2+b^2}}$

$\text{Taking reciprocals, we get}$

$\boxed{\bf\sec{\theta}=\dfrac{a^2+b^2}{2ab}}$

$\text{We know that,}$

$\tan{\theta}=\dfrac{\sin{\theta}}{\cos{\theta}}$

$\implies\tan{\theta}=\dfrac{\dfrac{a^2-b^2}{a^2+b^2}}{\dfrac{2ab}{a^2+b^2}}$

$\implies\boxed{\bf\tan{\theta}=\dfrac{a^2-b^2}{2ab}}$

$\text{Taking reciprocals, we get}$

$\boxed{\bf\cot{\theta}=\dfrac{2ab}{a^2-b^2}}$

Answer 2

Answer:

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