Question
Prove the following trigonometric identities.
$\frac{\tan^2\text{A}}{1+\tan^2\text{A}}+\frac{\cot^2\text{A}}{1+\cot^2\text{A}}=1$
$\frac{\tan^2\text{A}}{1+\tan^2\text{A}}+\frac{\cot^2\text{A}}{1+\cot^2\text{A}}=1$
✓
Answer
In the given question, we need to prove $\frac{\tan^2\text{A}}{1+\tan^2\text{A}}+\frac{\cot^2\text{A}}{1+\cot^2\text{A}}=1$
Here, we will first solve the L.H.S.
Now using $\tan\theta=\frac{\sin\theta}{\cos\theta}\text{ and }\cot\theta=\frac{\cos\theta}{\sin\theta},$ we get
$\text{L.H.S}=\frac{\tan^2\text{A}}{1+\tan^2\text{A}}+\frac{\cot^2\text{A}}{1+\cot^2\text{A}}$
$=\frac{\Big(\frac{\sin^2\text{A}}{\cos^2\text{A}}\Big)}{\Big(1+\frac{\sin^2\text{A}}{\cos^2\text{A}}\Big)}+\frac{\Big(\frac{\cos^2\text{A}}{\sin^2\text{A}}\Big)}{\Big(1+\frac{\cos^2\text{A}}{\sin^2\text{A}}\Big)}$
$=\frac{\Big(\frac{\sin^2\text{A}}{\cos^2\text{A}}\Big)}{\Big(\frac{\cos^2\text{A}+\sin^2\text{A}}{\cos^2\text{A}}\Big)}+\frac{\Big(\frac{\cos^2\text{A}}{\sin^2\text{A}}\Big)}{\Big(\frac{\sin^2\text{A}+\cos^2\text{A}}{\sin^2\text{A}}\Big)}$
$=\frac{\Big(\frac{\sin^2\text{A}}{\cos^2\text{A}}\Big)}{\Big(\frac{1}{\cos^2\text{A}}\Big)}+\frac{\Big(\frac{\cos^2\text{A}}{\sin^2\text{A}}\Big)}{\Big(\frac{1}{\sin^2\text{A}}\Big)} \ (\text{Using }\sin^2\theta+\cos^2\theta=1)$
On further solving by taking the reciprocal of the denominator, we get,
$=\frac{\Big(\frac{\sin^2\text{A}}{\cos^2\text{A}}\Big)}{\Big(\frac{1}{\cos^2\text{A}}\Big)}+\frac{\Big(\frac{\cos^2\text{A}}{\sin^2\text{A}}\Big)}{\Big(\frac{1}{\sin^2\text{A}}\Big)}$
$=\Big(\frac{\sin^2\text{A}}{\cos^2\text{A}}\Big)\Big(\frac{\cos^2\text{A}}{1}\Big)+\Big(\frac{\cos^2\text{A}}{\sin^2\text{A}}\Big)\Big(\frac{\sin^2\text{A}}{1}\Big)$
$=\sin^2\text{A}+\cos^2\text{A}\ (\text{using} \sin^2\theta+\cos^2\theta)$
$=1=\text{R.H.S}$
$\therefore\ \text{L.H.S}=\text{R.H.S}$
Here, we will first solve the L.H.S.
Now using $\tan\theta=\frac{\sin\theta}{\cos\theta}\text{ and }\cot\theta=\frac{\cos\theta}{\sin\theta},$ we get
$\text{L.H.S}=\frac{\tan^2\text{A}}{1+\tan^2\text{A}}+\frac{\cot^2\text{A}}{1+\cot^2\text{A}}$
$=\frac{\Big(\frac{\sin^2\text{A}}{\cos^2\text{A}}\Big)}{\Big(1+\frac{\sin^2\text{A}}{\cos^2\text{A}}\Big)}+\frac{\Big(\frac{\cos^2\text{A}}{\sin^2\text{A}}\Big)}{\Big(1+\frac{\cos^2\text{A}}{\sin^2\text{A}}\Big)}$
$=\frac{\Big(\frac{\sin^2\text{A}}{\cos^2\text{A}}\Big)}{\Big(\frac{\cos^2\text{A}+\sin^2\text{A}}{\cos^2\text{A}}\Big)}+\frac{\Big(\frac{\cos^2\text{A}}{\sin^2\text{A}}\Big)}{\Big(\frac{\sin^2\text{A}+\cos^2\text{A}}{\sin^2\text{A}}\Big)}$
$=\frac{\Big(\frac{\sin^2\text{A}}{\cos^2\text{A}}\Big)}{\Big(\frac{1}{\cos^2\text{A}}\Big)}+\frac{\Big(\frac{\cos^2\text{A}}{\sin^2\text{A}}\Big)}{\Big(\frac{1}{\sin^2\text{A}}\Big)} \ (\text{Using }\sin^2\theta+\cos^2\theta=1)$
On further solving by taking the reciprocal of the denominator, we get,
$=\frac{\Big(\frac{\sin^2\text{A}}{\cos^2\text{A}}\Big)}{\Big(\frac{1}{\cos^2\text{A}}\Big)}+\frac{\Big(\frac{\cos^2\text{A}}{\sin^2\text{A}}\Big)}{\Big(\frac{1}{\sin^2\text{A}}\Big)}$
$=\Big(\frac{\sin^2\text{A}}{\cos^2\text{A}}\Big)\Big(\frac{\cos^2\text{A}}{1}\Big)+\Big(\frac{\cos^2\text{A}}{\sin^2\text{A}}\Big)\Big(\frac{\sin^2\text{A}}{1}\Big)$
$=\sin^2\text{A}+\cos^2\text{A}\ (\text{using} \sin^2\theta+\cos^2\theta)$
$=1=\text{R.H.S}$
$\therefore\ \text{L.H.S}=\text{R.H.S}$
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