Question

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Answer 1

Question:-

$\sf \large Prove \: that:(1 + tanA \: tanB) {}^{2} + (tanA - tanB) {}^{2} = sec {}^{2} A \: sec {}^{2} B.$

Given:-

$\sf \large(1 + tanA \: tanB) {}^{2} + (tanA - tanB) {}^{2}.$

To Prove:-

$\sf \large(1 + tanA \: tanB) {}^{2} + (tanA - tanB) {}^{2} = sec {}^{2} A \: sec {}^{2} B.$

Solution:-

$\sf \large L.H.S. = (1 + tanA \: tanB) {}^{2} + (tanA - tanB) {}^{2}.$

$\sf \large \Rightarrow1 + tan {}^{2}A \: tan {}^{2} B + 2 \: tanA \: tanB + tan {}^{2} A + tan {}^{2} B - 2 \: tanA \: tanB.$

$\sf \large \Rightarrow = (1 + tan {}^{2} A) + tan {}^{2} A \: tan {}^{2} B + tan {}^{2} B.$

$\sf \large \Rightarrow sec {}^{2} A + tan{}^{2} B(tan {}^{2} A + 1) .\: \: \: \: \: \: ( \because sec {}^{2} A - tan {}^{2} A = 1).$

$\sf \large \Rightarrow sec {}^{2} A + sec {}^{2} A \: \: tan {}^{2} B.$

$\sf \large \Rightarrow sec {}^{2} A(1 + tan {}^{2} B). \: \: \: \: \: \: \: \: ( \because sec {}^{2} B - tan {}^{2}B = 1).$

$\sf \large \Rightarrow \: \: R.H.S. = sec {}^{2} A \: sec {}^{2} B.$

Answer:-

${ \boxed{ \sf \large \red{Hence \: Proved, L.H.S. = R.H.S.}}} \\ { \boxed{ \sf \large \blue{(1 + tanA \: tanB) {}^{2} + (tanA - tanB) {}^{2} = sec {}^{2} A \: sec {}^{2} B.}}}$

Hope you have satisfied. ⚘