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Trigonometric Identities — Maths STD 10 — Question

Maharashtra BoardEnglish MediumSTD 10MathsTrigonometric Identities3 Marks
Question
Prove the following trigonometric identities.
If $\text{x}=\text{a}\sec\theta+\text{b}\tan\theta\text{ and y}=\text{a}\tan\theta+\text{b}\sec\theta,$ prove that $x^2 - y^2 = a^2 - b^2$.
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Answer

$\text{L.H.S}=\text{x}^2-\text{y}^2$
$=(\text{a}\sec\theta+\text{b}\tan\theta)^2-(\text{a}\tan\theta+\text{b}\sec\theta)^2$
$=\text{a}^2\sec^2+\text{b}^2\tan^2\theta+2\text{ab}\sec\theta\tan\theta$
$=-\text{a}^2\tan^2\theta-\text{b}^2\sec\theta-2\text{ab}\sec\theta\tan\theta$
$=\text{a}^2-\sec^2\theta-\text{b}^2\sec^2+\text{b}^2\tan^2\theta-\text{a}^2\tan^2\theta$
$=\sec^2\theta(\text{a}^2-\text{b}^2)+\tan^2\theta(\text{b}^2-\text{a}^2)$
$=\sec^2\theta(\text{a}^2-\text{b}^2)-\tan^2\theta(\text{a}^2-\text{b}^2)$
$=(\text{a}^2-\text{b}^2)(\sec^2\theta-\tan^2\theta) \big[\because\ \sec^2\theta-\tan^2\theta=1\big]$
$=\text{a}^2-\text{b}^2=\text{R.H.S}$
$\therefore\ \text{L.H.S}=\text{R.H.S}$

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