Question

If x=b sec alpha +a tan alpha and y =a sec alpha +b tan alpha prove that x^2-y^2=b^2-a^2

Answer 1

❏ To ProvE

$If, \bf x=b\sec\alpha+a\tan\alpha\\</p><p> \bf y=a\sec\alpha+b\tan\alpha\\</p><p> then\:prove \:that \:\\</p><p> \bf \boxed{x{}^{2}-y{}^{2}=b{}^{2}-a{}^{2}}\\</p><p>$

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❏ ProvinG:-

$L.H.S.= \bf x{}^{2}-y{}^{2}\\= {b\sec\alpha+a\tan\alpha}^{2}-{a\sec\alpha+b\tan\alpha}^{2}\\</p><p> =b{}^{2}{\sec\alpha}^{2}+a{}^{2}{\tan\alpha}^{2}+2(b\sec\alpha )(a\tan\alpha)\\ \:\:\:-\{a{}^{2}{\sec}^{2}\alpha+b{}^{2}{\tan}^{2}\alpha+2(a\sec\alpha)(b\tan\alpha)\}\\</p><p> =b{}^{2}{\sec}^{2}\alpha+a{}^{2}{\tan}^{2}\alpha+\cancel{2ab\sec\alpha\tan\alpha}\\ \:\:\:-a{}^{2}{\sec}^{2}\alpha-b{}^{2}{\tan}^{2}\alpha-\cancel{2ab\sec\alpha\tan\alpha}\\</p><p>=b{}^{2}[{\sec}^{2}\alpha-{\tan}^{2}\alpha]-a{}^{2}[{\sec}^{2}\alpha-{\tan}^{2}\alpha]\\</p><p>=b{}^{2}[1]-a{}^{2}[1]\\</p><p>=\boxed{\bf \red{b{}^{2}-a{}^{2}=R.H.S.\:(proved)}}$

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$\mathbb{\red{SOME \:TRIGONOMETRIC\: IDENTITIES}}$

$\begin{cases}</p><p> \Longrightarrow \sin^{2}\theta+\cos^{2}\theta=1\\</p><p></p><p>\bf\Longrightarrow \sec^{2}\theta-\tan^{2}\theta=1\\</p><p></p><p>\bf\Longrightarrow \cosec^{2}\theta-\cot^{2}\theta=1\\</p><p>\end{cases}$

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$\underline{ \huge\mathfrak{hope \: this \: helps \: you}}$

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