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Trigonometric Identities — Maths STD 10 — Question

Maharashtra BoardEnglish MediumSTD 10MathsTrigonometric Identities4 Marks
Question
Prove the following trigonometric identities.
$(\text{cosec }\theta-\sec\theta)(\cot\theta-\tan\theta)=(\text{cosec }\theta+\sec\theta)(\sec\theta\text{ cosec }\theta-2)$
✓

Answer

We have to prove $(\text{cosec }\theta-\sec\theta)(\cot\theta-\tan\theta)=(\text{cosec }\theta+\sec\theta)(\sec\theta\text{ cosec }\theta-2)$
We know that, $\sin^2\theta+\cos^2\theta=1$
Consider the L.H.S
$\text{L.H.S}=(\text{cosec }\theta-\sec\theta)(\cot\theta-\tan\theta)$
$=\Big(\frac{1}{\sin\theta}-\frac{1}{\cos\theta}\Big)\Big(\frac{\cos\theta}{\sin\theta}-\frac{\sin\theta}{\cos\theta}\Big)$
$=\Big(\frac{\cos\theta-\sin\theta}{\sin\theta\cos\theta}\Big)\Big(\frac{\cos^2\theta-\sin^2\theta}{\sin\theta\cos\theta}\Big)$
$=\frac{(\cos\theta-\sin\theta)}{\sin\theta\cos\theta}\frac{(\cos\theta+\sin\theta)(\cos\theta-\sin\theta)}{\sin\theta\cos\theta}$
$=\frac{(\cos\theta+\sin\theta)(\cos\theta-\sin\theta)^2}{\sin^2\theta\cos^2\theta}$
Now, consider the R.H.S
$\text{R.H.S}=(\text{cosec }\theta+\sec\theta)(\sec\theta\text{ cosec }\theta-2)$
$=\Big(\frac{1}{\sin\theta}+\frac{1}{\cos\theta}\Big)\Big(\frac{1}{\cos\theta}\frac{1}{\sin\theta}-2\Big)$
$=\Big(\frac{\cos\theta+\sin\theta}{\sin\theta\cos\theta}\Big)\Big(\frac{1-2\sin\theta\cos\theta}{\sin\theta\cos\theta}\Big)$
$=\frac{(\cos\theta+\sin\theta)}{\sin\theta\cos\theta}\frac{(\cos^2\theta+\sin^2\theta-2\cos\theta\sin\theta)}{\sin\theta\cos\theta}$
$=\frac{(\cos\theta+\sin\theta)(\cos\theta-\sin\theta)^2}{\sin^2\theta\cos^2\theta}$
$\therefore \text{L.H.S}=\text{R.H.S}$

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