Prove the following trigonometric identities.
Question:

Prove the following trigonometric identities.

If $\operatorname{cosec} \theta-\sin \theta=a^{3}, \sec \theta-\cos \theta=b^{3}$, prove that $a^{2} b^{2}\left(a^{2}+b^{2}\right)=1$

Solution:

Given that,

$\operatorname{cosec} \theta-\sin \theta=a^{3}$…….(1)

 

$\sec \theta-\cos \theta=b^{3}$……..(2)

We have to prove $a^{2} b^{2}\left(a^{2}+b^{2}\right)=1$

 

We know that $\sin ^{2} \theta+\cos ^{2} \theta=1$

Now from the first equation, we have

$\operatorname{cosec} \theta-\sin \theta=a^{3}$

$\Rightarrow \frac{1}{\sin \theta}-\sin \theta=a^{3}$

$\Rightarrow \quad \frac{1-\sin ^{2} \theta}{\sin \theta}=a^{3}$

$\Rightarrow \quad \frac{\cos ^{2} \theta}{\sin \theta}=a^{3}$

$\Rightarrow \quad \frac{\cos ^{2} \theta}{\sin \theta}=a^{3}$

$\Rightarrow \quad a=\frac{\cos ^{1 / 3} \theta}{\sin ^{1 / 3} \theta}$

Again from the second equation, we have

$\sec \theta-\cos \theta=b^{3}$

$\Rightarrow \frac{1}{\cos \theta}-\cos \theta=b^{3}$

$\Rightarrow \quad \frac{1-\cos ^{2} \theta}{\cos \theta}=b^{3}$

$\Rightarrow \quad \frac{\sin ^{2} \theta}{\cos \theta}=b^{3}$

$\Rightarrow \quad b=\frac{\sin ^{2 / 3} \theta}{\cos ^{1 / 3} \theta}$

Therefore, we have

$a^{2} b^{2}\left(a^{2}+b^{2}\right)=\frac{\cos ^{4 / 3} \theta}{\sin ^{2 / 3} \theta} \frac{\sin ^{4 / 3} \theta}{\cos ^{2 / 3} \theta}\left(\frac{\cos ^{4 / 3} \theta}{\sin ^{2 / 3} \theta}+\frac{\sin ^{4 / 3} \theta}{\cos ^{2 / 3} \theta}\right)$

$=\sin ^{2 / 3} \theta \cos ^{2 / 3} \theta\left(\frac{\cos ^{4 / 3} \theta}{\sin ^{2 / 3} \theta}+\frac{\sin ^{4 / 3} \theta}{\cos ^{2 / 3} \theta}\right)$

$=\cos ^{2 / 3} \theta \cos ^{4 / 3} \theta+\sin ^{2 / 3} \theta \sin ^{4 / 3} \theta$

$=\cos ^{2} \theta+\sin ^{2} \theta$

 

$=1$

Hence proved.

 

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