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Trigonometric Identities — Maths STD 10 — Question

Maharashtra BoardEnglish MediumSTD 10MathsTrigonometric Identities4 Marks
Question
Prove the following trigonometric identities.
$(1+\tan^2\text{A})+\Big(1+\frac{1}{\tan^2\text{A}}\Big)=\frac{1}{\sin^2\text{A}-\sin^4\text{A}}$
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Answer

We need to prove $(1+\tan^2\text{A})+\Big(1+\frac{1}{\tan^2\text{A}}\Big)=\frac{1}{\sin^2\text{A}-\sin^4\text{A}}$
Using the property $1+\tan^2\theta=\sec^2\theta,$ we get
$\text{L.H.S}=(1+\tan^2\text{A})+\Big(1+\frac{1}{\tan^2\text{A}}\Big)=\sec^2\text{A}+\Big(\frac{\tan^2\text{A}+1}{\tan^2\text{A}}\Big)$
$=\sec^2\text{A}+\Big(\frac{\sec^2\text{A}}{\tan^2\text{A}}\Big)$
Now, using $\sec\theta=\frac{1}{\cos\theta}\text{ and }\tan\theta=\frac{\sin\theta}{\cos\theta},$ we get
$\sec^2\text{A}+\Big(\frac{\sec^2\text{A}}{\tan^2\text{A}}\Big)=\frac{1}{\cos^2\text{A}}+\Bigg(\frac{\frac{1}{\cos^2\text{A}}}{\frac{\sin^2\text{A}}{\cos^2\text{A}}}\Bigg)$
$=\frac{1}{\cos^2\text{A}}+\Big(\frac{1}{\cos^2\text{A}}\times\frac{\cos^2\text{A}}{\sin^2\text{A}}\Big)$
$=\frac{1}{\cos^2\text{A}}+\frac{1}{\sin^2\text{A}}$
$=\frac{\sin^2\text{A}+\cos^2\text{A}}{\cos^2\text{A}(\sin^2\text{A})}$
Further, using the peroperty, $\sin^2\theta+\cos^2\theta=1,$ we get
$\frac{\sin^2\text{A}+\cos^2\text{A}}{\cos^2\text{A}(\sin^2\text{A})}=\frac{1}{\cos^2\text{A}(\sin^2\text{A})}$
$=\frac{1}{(1-\sin^2\text{A})(\sin^2\text{A})}\ (\text{using}\cos^2\theta=1-\sin^2\theta)$
$=\frac{1}{\sin^2\text{A}-\sin^4\text{A}}=\text{R.H.S}$
$\therefore\ \text{L.H.S}=\text{R.H.S}$

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