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Trigonometric Identities — Maths STD 10 — Question

Maharashtra BoardEnglish MediumSTD 10MathsTrigonometric Identities5 Marks
Question
If $\text{x}=\text{a}\sec\theta+\text{b}\tan\theta$ and $\text{y}=\text{a}\tan\theta-\text{b}\sec\theta,$ prove that $\big(\text{x}^2+\text{y}^2\big)=\big(\text{a}^2-\text{b}^2\big).$
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Answer

We have $\text{x}^2+\text{y}^2=\big[(\text{a}\sec\theta+\text{b}\tan\theta)^2+(\text{a}\tan\theta-\text{b}\sec\theta)^2\big]$
$=\big(\text{a}^2\sec^2\theta+\text{b}^2\tan^2\theta+2\text{ab}\sec\theta\tan\theta\big)\\ \ -\big(\text{a}^2\tan^2\theta+\text{b}^2\sec^2\theta+2\text{ab}\tan\theta\sec\theta\big)$
$=\text{a}^2\sec^2\theta+\text{b}^2\tan^2\theta-\text{a}^2\tan^2\theta+\text{b}^2\sec^2\theta$
$=\big(\text{a}^2\sec^2\theta+\text{a}^2\tan^2\theta\big)-\big(\text{b}^2\sec^2\theta+\text{b}^2\tan^2\theta\big)$
$=\text{a}^2\big(\sec^2\theta-\tan^2\theta\big)-\text{b}^2\big(\sec^2\theta-\tan^2\theta\big)$
$=\text{a}^2-\text{b}^2$ $\big[\because\ \sec^2+\tan^2=1\big]$
Hence, $\text{x}^2-\text{y}^2=\text{a}^2-\text{b}^2$

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