If = a^2-b^2 a^2+b^2 find the value of , and cosec x

Question

If $\sin\text{x}=\frac{\text{a}^2-\text{b}^2}{\text{a}^2+\text{b}^2}$ find the value of $\tan\text{x},\sec\text{x}$ and $\text{cosec x}$

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Answer

Now, $\cos\text{x}=\sqrt{1-\sin^2\text{x}}$ $=\sqrt{\frac{1-(\text{a}^2-\text{b}^2)}{(\text{a}^2+\text{b}^2)^2}}$ $\Big[\because\sin\text{x}=\frac{\text{a}^2-\text{b}^2}{\text{a}^2+\text{b}^2}\Big]$$=\sqrt{\frac{(\text{a}^2+\text{b}^2)^2-(\text{a}^2-\text{b}^2)^2}{(\text{a}^2+\text{b}^2)^2}}$
$=\sqrt{\frac{(\text{a}^2+\text{b}^2+\text{a}^2-\text{b}^2)(\text{a}^2+\text{b}^2-\text{a}^2+\text{b}^2)}{\text{a}^2+\text{b}^2}}$ $(\text{Using x}^2-\text{y}^2=(\text{x}-\text{y})(\text{x+y}))$ $=\sqrt{\frac{2\text{a}^2\times2\text{b}^2}{\text{a}^2+\text{b}^2}}$ $=\frac{2\text{ab}}{\text{a}^2+\text{b}^2}\cdots\text{(ii)}$ Now, $\tan\text{x}=\frac{\sin\text{x}}{\cos\text{x}}$ $=\frac{\text{a}^2-\text{b}^2}{\frac{\text{a}^2+\text{b}^2}{\frac{2\text{ab}}{\text{a}^2+\text{b}^2}}}$ $=\frac{\text{a}^2-\text{b}^2}{2\text{}ab}$ $\sec=\text{x}\frac{1}{\cos\text{x}}=\frac{\text{a}^2+\text{b}^2}{2\text{ab}}$ (From(ii)) and $\text{cosec}\text{ x}=\frac{1}{\sin\text{x}}=\frac{\text{a}^2+\text{b}^2}{\text{a}^2-\text{b}^2}$ (From(i))

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