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Trigonometric Identities — Maths STD 10 — Question

Maharashtra BoardEnglish MediumSTD 10MathsTrigonometric Identities4 Marks
Question
Prove the following trigonometric identities.
If $\text{cosec }\theta-\sin\theta=\text{a}^3,\sec\theta-\cos\theta=\text{b}^3,$ prove that $a^2b^2 (a^2 + b^2) = 1$.
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Answer

$\text{cosec }\theta-\sin\theta=\text{a}^3$
$\frac{1}{\sin\theta}-\sin\theta=\text{a}^3$
$\frac{1-\sin^2}{\sin\theta}=\text{a}^3$
$\frac{\cos^2\theta}{\sin\theta}=\text{a}^3$
$\text{a}=\frac{\cos^{\frac{2}{3}}\theta}{\sin^{\frac{1}{3}}\theta}$
$\Rightarrow\ \text{a}^2=\frac{\cos\frac{4}{3}\theta}{\sin\frac{2}{3}\theta}$
$\sec\theta-\cos\theta=\text{B}^3$
$\frac{1}{\cos\theta}-\cos\theta=\text{b}^3$
$\frac{1-\cos^2\theta}{\cos\theta}=\text{b}^3$
$\frac{\sin^2\theta}{\cos\theta}=\text{b}^3$
$\text{b}=\frac{\sin\frac{2}{3}\theta}{\cos\frac{1}{3}\theta}$
Now, $\text{L.H.S}=\text{a}^2\text{b}^2(\text{a}^2+\text{b}^2)$
$=\frac{\cos^{\frac{4}{3}}\theta}{\sin^{\frac{2}{3}}\theta}\times\frac{\sin^\frac{4}{3}\theta}{\cos^\frac{2}{3}\theta}\bigg(\frac{\cos^{\frac{4}{3}}\theta}{\sin^{\frac{2}{3}\theta}}+\frac{\sin^{\frac{4}{3}}\theta}{\cos^{\frac{2}{3}}\theta}\bigg)$
$=\cos^{\frac{4}{3}-\frac{2}{3}}\theta\sin^{\frac{4-2}{3}}\bigg(\frac{\cos^{\frac{4}{3}}\theta}{\sin^{\frac{2}{3}\theta}}+\frac{\sin^{\frac{4}{3}}\theta}{\cos^{\frac{2}{3}}\theta}\bigg)$
$=\cos^{\frac{2}{3}}\theta\sin^\frac{2}{3}\bigg(\frac{1}{\sin^{\frac{2}{3}}\theta\cos^{\frac{2}{3}}\theta}\bigg) \big(\because \cos^2\theta+\sin^2\theta=1\big)$
$=1=\text{R.H.S}$
$\therefore\ \text{L.H.S}=\text{R.H.S}$

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