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Trigonometric Functions — MATHS STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceMATHSTrigonometric Functions5 Marks
Question
Prove the following identities:
$(1+\tan\alpha\tan\beta)^2+(\tan\alpha-\tan\beta)^2=\sec^2\alpha\sec^2\beta$
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Answer

$\text{L.H.S}=(1+\tan\alpha\tan\beta)^2+(\tan\alpha-\tan\beta)^2$
$=1(\tan\alpha+\tan\beta)^2+2.1\tan\alpha\tan\beta\\\ \ +(\tan\alpha)^2+(\tan\beta)^2-1\tan\alpha.\tan\beta$ $(\text{Using(a+b)}^2=\text{a}^2+\text{b}^2+2\text{ab and (a-b)}^2=\text{a}^2+\text{b}^2-2\text{ab})$
$=1+\tan^2\alpha+\tan^2\beta+2\tan\alpha\tan\beta+\tan^2\alpha+\tan^2\beta-2\tan\alpha\tan\beta$
$=1+\tan^2\alpha+\tan^2\alpha+\tan^2\beta+\tan^2\beta$
$=\sec^2\alpha+\tan^2\beta(1+\tan^2\alpha)$$(\because1+\tan^2\alpha=\sec^2\alpha)$
$=\sec^2\alpha+\tan^2\beta.\sec^2\alpha$
$=\sec^2\alpha(1+\tan^2\beta)$
$=\sec^2\alpha.\sec^2\beta$ $(\because1+\tan^2\beta=\sec^2\beta)$
$=\text{R.H.S}$
$\text{Proved}$

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