If theta = 3alpha and sin theta = frac{a}{sqr | vedclass

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(A) Given $	heta = 3\alpha$ and $\sin 3\alpha = \frac{a}{\sqrt{a^2 + b^2}}$.Let $\cos 3\alpha = \frac{b}{\sqrt{a^2 + b^2}}$.We need to evaluate $E =

Vedclass · Accepted Answer

$2\sqrt{a^2 + b^2}$

Vedclass · Answer

(A) Given $	heta = 3\alpha$ and $\sin 3\alpha = \frac{a}{\sqrt{a^2 + b^2}}$.Let $\cos 3\alpha = \frac{b}{\sqrt{a^2 + b^2}}$.We need to evaluate $E = a \csc \alpha - b \sec \alpha = \frac{a}{\sin \alpha} - \frac{b}{\cos \alpha} = \frac{a \cos \alpha - b \sin \alpha}{\sin \alpha \cos \alpha}$.Substitute $a = \sqrt{a^2 + b^2} \sin 3\alpha$ and $b = \sqrt{a^2 + b^2} \cos 3\alpha$:$E = \frac{\sqrt{a^2 + b^2} (\sin 3\alpha \cos \alpha - \cos 3\alpha \sin \alpha)}{\sin \alpha \cos \alpha}$.Using the identity $\sin(A - B) = \sin A \cos B - \cos A \sin B$:$E = \frac{\sqrt{a^2 + b^2} \sin(3\alpha - \alpha)}{\sin \alpha \cos \alpha} = \frac{\sqrt{a^2 + b^2} \sin 2\alpha}{\sin \alpha \cos \alpha}$.Since $\sin 2\alpha = 2 \sin \alpha \cos \alpha$:$E = \frac{\sqrt{a^2 + b^2} (2 \sin \alpha \cos \alpha)}{\sin \alpha \cos \alpha} = 2\sqrt{a^2 + b^2}$.

If $\theta = 3\alpha$ and $\sin \theta = \frac{a}{\sqrt{a^2 + b^2}}$,then the value of the expression $a \csc \alpha - b \sec \alpha$ is:

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