If cosec α = √2, evaluate $$\frac{(2 sin²⁡α+3 cos²⁡α)}{(cosec^2 α+cot^2 α)}$$​​

Correct option is A

Given:

​$$cosec \alpha = \sqrt{2}$$​​

Formula used:

$$cosec \alpha = \frac{1}{sin \alpha}$$​

​$$\sin \alpha = 1/\sqrt{2}$$​​

$$\sin^2 \alpha + \cos^2 \alpha = 1$$​​

​$$\cos \alpha = \sqrt{1 - \sin^2 \alpha}$$​​

Solution:

​$$\sin \alpha = \frac{1}{\sqrt{2}} , so \sin^2 \alpha =\frac{ 1}{2}$$​​

​$$\cos^2 \alpha = 1 -\frac{ 1}{2}= \frac{1}{2}$$​​

​$$\frac{2\sin^2 \alpha + 3\cos \alpha}{\cosec^2 \alpha + \cot^2 \alpha}$$​​

First, calculate the numerator:

Numerator = $$2\sin^2 \alpha + 3\cos^2 \alpha$$​​

Numerator = $$2\times\frac{1}{2} + 3\times\frac{1}{{2}}$$​​

Numerator = $$1 + \frac{3}{{2}} = 2.5$$​​

Now, calculate the denominator:

Denominator = $$\cosec^2 \alpha + \cot^2 \alpha$$​​

We know $$cosec \alpha = \sqrt{2}$$​​

​$$\cosec^2 \alpha = (\sqrt{2})^2 = 2$$​​

​$$\cot \alpha = \frac{cos \alpha }{\sin \alpha} =\frac{\frac{ 1}{\sqrt{2}}} { \frac{1}{\sqrt{2}}}= 1$$​​

​$$\cot^2 \alpha = 1$$​​

Denominator = 2 + 1 = 3

Finally, the expression becomes:

​$$\text{Result} = \frac{2.5}{3} = \frac{5}{6}$$​​