Question

If $2{\sec }^{2}A-{\sec }^{4}A-2{\text{cosec}}^{2}A+{\text{cosec}}^{4}A=\frac{15}{4}$, then $\tan A$ is equal to

• A. $\frac{1}{\sqrt{2}}$
• B. $\frac{1}{2}$
• C. $\frac{1}{2\sqrt{2}}$
• D. $\frac{-1}{\sqrt{2}}$

Answer 1

Answer: (A), (D)

The correct options are
A $\frac{1}{\sqrt{2}}$
D $\frac{-1}{\sqrt{2}}$

$2{\sec }^{2}A-{\sec }^{4}A-2{\csc }^{2}A+{\csc }^{4}A=\frac{15}{4}\Rightarrow 2(1+{\tan }^{2}A)-{(1+{\tan }^{2}A)}^2-2(1+{\cot }^{2}A)+{(1+{\cot }^{2}A)}^2=\frac{15}{4}\Rightarrow 2+2{\tan }^{2}A-1-{\tan }^{4}A-2{\tan }^{2}A-2-2{\cot }^{2}A+1+{\cot }^{4}A+2{\cot }^{2}A=\frac{15}{4}\Rightarrow {\cot }^{4}A-{\tan }^{4}A=\frac{15}{4}$
Let ${\tan }^{4}A=y$
$\frac{1}{y}-y=\frac{15}{4}\Rightarrow 4y^2+15y-4=0\therefore y=-4,\frac{1}{4}$
$\Rightarrow {\tan }^{4}A\ne -4\{\because {\tan }^{4}A>0\}$
And ${\tan }^{4}A=\frac{1}{4}\therefore \tan A=\pm \frac{1}{\sqrt{2}}$

Hence option $A$ and $D$ both are correct.