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If $ \operatorname{cosec} \theta-\sin \theta=a^{3}, \sec \theta-\cos \theta=b^{3} $, prove that $ a^{2} b^{2}\left(a^{2}+b^{2}\right)=1 $
Given:
\( \operatorname{cosec} \theta-\sin \theta=a^{3}, \sec \theta-\cos \theta=b^{3} \)
To do:
We have to prove that \( a^{2} b^{2}\left(a^{2}+b^{2}\right)=1 \).
Solution:
We know that,
$\sin^2 A+\cos^2 A=1$
$\operatorname{cosec}^2 A-\cot^2 A=1$
$\sec^2 A-\tan^2 A=1$
$\cot A=\frac{\cos A}{\sin A}$
$\tan A=\frac{\sin A}{\cos A}$
$\operatorname{cosec} A=\frac{1}{\sin A}$
$\sec A=\frac{1}{\cos A}$
Therefore,
$\operatorname{cosec} \theta-\sin \theta=a^{3}$
$\Rightarrow \frac{1}{\sin \theta}-\sin \theta=a^{3}$
$\Rightarrow \frac{1-\sin ^{2} \theta}{\sin \theta}=a^{3}$
$\Rightarrow \frac{\cos ^{2} \theta}{\sin \theta}=a^{3}$
$\Rightarrow \left(\frac{\cos ^{2} \theta}{\sin \theta}\right)^{\frac{1}{3}}=a$
$\sec \theta-\cos \theta=b^{3}$
$\Rightarrow \frac{1}{\cos \theta}-\cos \theta=b^{3}$
$\Rightarrow \frac{1-\cos ^{2} \theta}{\cos \theta}=b^{3}$
$\Rightarrow \frac{\sin ^{2} \theta}{\cos \theta}=b^{3}$
$\Rightarrow \left(\frac{\sin ^{2} \theta}{\cos \theta}\right)^{\frac{1}{3}}=b$
This implies,
$a^{2} b^{2}(a^{2}+b^{2})=(\frac{\cos ^{2} \theta}{\sin\theta})^{\frac{2}{3}}(\frac{\sin ^{2} \theta}{\cos \theta})^{\frac{2}{3}}[(\frac{\cos ^{2} \theta}{\sin \theta})^{\frac{2}{3}}+(\frac{\sin ^{2} \theta}{\cos \theta})^{\frac{2}{3}}]$
$=\left[\left(\frac{\cos ^{2} \theta}{\sin \theta}\right) \times\left(\frac{\sin ^{2} \theta}{\cos \theta}\right)\right]^{\frac{2}{3}}\left[\frac{\cos \theta^{\frac{4}{3}}}{\sin \theta^{\frac{2}{3}}}+\frac{\sin \theta^{\frac{4}{3}}}{\cos \theta^{\frac{2}{3}}}\right]$
$=(\sin \theta \cos \theta)^{\frac{2}{3}}\left[\frac{\cos \theta^{\frac{6}{3}}+\sin \theta^{\frac{6}{3}}}{\sin \theta^{\frac{2}{3}} \cos \theta^{\frac{2}{3}}}\right]$
$=\frac{\sin \theta^{\frac{2}{3}} \cos \theta^{\frac{2}{3}}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)}{\sin \theta^{\frac{2}{3}} \cos \theta^{\frac{2}{3}}}$
$=\cos ^{2} \theta+\sin ^{2} \theta$
$=1$
Hence proved.
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