Question
If $\text{cosec A}=\sqrt{2},$ find the value of $\frac{2\sin^2\text{A}+3\cot^2\text{A}}{4(\tan^2-\cos^2\text{A})}$.
✓
Answer
We know that $\cot\text{A}=\sqrt{\text{cosec}^2\text{A}-1}$
$=\sqrt{(2)^2-1}=\sqrt{2-1}$
$=-1$
$\tan\text{A}=\frac{1}{\text{cot}A}=\frac{1}{1}=1$
$\sin\text{A}=\frac{1}{\text{cosec}}\text{A}=\frac{1}{\sqrt{2}}\therefore\sin\text{A}=\frac{1}{\sqrt{2}}$
$\cos\text{A}=\sqrt{1-\sin^2\text{A}}$
$=\sqrt{1-\Big(\frac{1}{\sqrt{2}}\Big)^2}-\sqrt{\frac{1}{\sqrt{2}}}=\sqrt{\frac{1}{\sqrt{2}}}.$
On substituting we get
$\frac{2\Big[\frac{1}{\sqrt{2}}\Big]^3+3\big[1\big]^2}{4\Big[(1)-\big(\frac{1}{\sqrt{2}}\big)^2\Big]}=\frac{2=\frac{1}{2}+3}{4\Big[1-\frac{1}{2}\Big]}$
$\Rightarrow\frac{1+3}{4\times\frac{1}{2}}=\frac{4}{2}=2$
$=\sqrt{(2)^2-1}=\sqrt{2-1}$
$=-1$
$\tan\text{A}=\frac{1}{\text{cot}A}=\frac{1}{1}=1$
$\sin\text{A}=\frac{1}{\text{cosec}}\text{A}=\frac{1}{\sqrt{2}}\therefore\sin\text{A}=\frac{1}{\sqrt{2}}$
$\cos\text{A}=\sqrt{1-\sin^2\text{A}}$
$=\sqrt{1-\Big(\frac{1}{\sqrt{2}}\Big)^2}-\sqrt{\frac{1}{\sqrt{2}}}=\sqrt{\frac{1}{\sqrt{2}}}.$
On substituting we get
$\frac{2\Big[\frac{1}{\sqrt{2}}\Big]^3+3\big[1\big]^2}{4\Big[(1)-\big(\frac{1}{\sqrt{2}}\big)^2\Big]}=\frac{2=\frac{1}{2}+3}{4\Big[1-\frac{1}{2}\Big]}$
$\Rightarrow\frac{1+3}{4\times\frac{1}{2}}=\frac{4}{2}=2$
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