Question

Plss answer ......if sec theta × cos alpha =root 2 tan theta × cot alpha =root 3 find the value of sin theta and sin alpha

Answer 1

Answer: The answer is $\sin \theta=\pm\dfrac{1}{\sqrt2}$ and $\sin \alpha=\pm\dfrac{\sqrt3}{2}.$

Step-by-step explanation: We are given that

$\sec \theta\cos \alpha=\sqrt2~~~\textup{and}~~~\tan \theta\cot \alpha=\sqrt3.$

To find the value of $\sin \theta~~\textup{and}~~\sin \alpha.$

Now, dividing the first equation by the second, we have

$\dfrac{\sec \theta \cos \alpha}{\tan \theta \cot \alpha}=\dfrac{\sqrt2}{\sqrt3}\\\\\\\Rightarrow \dfrac{\cos \alpha}{\cos \theta}\times \dfrac{\cos \theta \sin \alpha}{\sin \theta \cos \alpha}=\dfrac{\sqrt2}{\sqrt3}\\\\\\\Rightarrow \dfrac{\sin \alpha}{\sin \theta}=\dfrac{\sqrt2}{\sqrt3}\\\\\\\Rightarrow \dfrac{\sin \alpha}{\sqrt2}=\dfrac{\sin \theta}{\sqrt3}=k(let).$

∴$\sin \alpha=k\sqrt2~~~\textup{and}~~~\sin \theta=k\sqrt3.$

Again, multiplying both the equations, we have

$\sec \theta\cos \alpha\times \tan \theta\cot \alpha=\sqrt6\\\\\Rightarrow \dfrac{\cos \alpha}{\cos \theta}\times \dfrac{\sin \theta\cos \alpha}{\cos \theta\sin \alpha}=\sqrt6\\\\\\\Rightarrow \dfrac{\cos^2 \alpha}{\cos^2 \theta}\times \dfrac{\sqrt3}{\sqrt2}=\sqrt6\\\\\\\Rightarrow \dfrac{\cos^2 \alpha}{\cos^2 \theta}=2\\\\\\\Rightarrow \dfrac{1-\sin^2 \alpha}{1-\sin^2 \theta}=2\\\\\\\Rightarrow 1-2k^2=2-6k^2\\\\\Rightarrow k^2=\dfrac{1}{4}\\\\\\\Rightarrow k=\pm\dfrac{1}{2}.$

Thus, the required values are

$\sin \alpha=\pm\dfrac{1}{\sqrt2}~~\textup{and}~~\sin \theta=\pm\dfrac{\sqrt3}{2}.$

Answer 2

sin alpha = 1/root 2.

sin theta = root 3/2