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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