If $ \operatorname{cosec} A=\sqrt{2} $, find the value of $ \frac{2 \sin ^{2} A+3 \cot ^{2} A}{4\left(\tan ^{2} A-\cos ^{2} A\right)} $

Given:

\( \operatorname{cosec} A=\sqrt{2} \)

To do:

We have to find the value of \( \frac{2 \sin ^{2} A+3 \cot ^{2} A}{4\left(\tan ^{2} A-\cos ^{2} A\right)} \).

Solution:  

$\operatorname{cosec} A=\sqrt{2}$

$\Rightarrow \operatorname{cosec}^{2} \mathrm{~A}=(\sqrt{2})^2$

$\Rightarrow \operatorname{cosec}^{2} \mathrm{~A}=2$

$\sin ^{2} \mathrm{~A}=\frac{1}{\operatorname{cosec}^{2} \mathrm{~A}}$

$=\frac{1}{2}$

$\cos ^{2} \mathrm{~A}=1-\sin ^{2} \mathrm{~A}$

$=1-\frac{1}{2}$

$=\frac{1}{2}$

$\tan ^{2} \mathrm{~A}=\frac{\sin ^{2} \mathrm{~A}}{\cos ^{2} \mathrm{~A}}$

$=\frac{\frac{1}{2}}{\frac{1}{2}}$

$=1$

$\cot ^{2} \mathrm{~A}=\frac{1}{\tan ^{2} \mathrm{~A}}$

$=1$

Therefore,

$\frac{2 \sin ^{2} \mathrm{~A}+3 \cot ^{2} \mathrm{~A}}{4(\tan ^{2} \mathrm{~A}-\cos ^{2} \mathrm{~A})}=\frac{2 \times \frac{1}{2}+3 \times 1}{4(1-\frac{1}{2})}$

$=\frac{1+3}{4 \times \frac{1}{2}}$

$=\frac{4}{2}$

$=2$

The value of \( \frac{2 \sin ^{2} A+3 \cot ^{2} A}{4\left(\tan ^{2} A-\cos ^{2} A\right)} \) is $2$.

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Updated on: 10-Oct-2022

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