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Sum of Trigonometric Ratios in Terms of Their Product

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Question

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$(sec A + cos A) (sec A - cos A) = {tan}^{2} A + {sin}^{2} A$

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Solution

$(sec A + cos A) (sec A - cos A)$
$= {sec}^{2} A - {cos}^{2} A$
$= 1 + {tan}^{2} A - 1 + {sin}^{2} A = {tan}^{2} A + {sin}^{2} A$

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Sum of Trigonometric Ratios in Terms of Their Product

Standard XII Mathematics

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