Question

Given $a={\sec }^2\alpha ,b={cosec}^2\alpha ,and,c=\frac{1}{1-{\sin }^2\alpha {\cos }^2\alpha }.$ Find the value of $\frac{1}{a^2}+\frac{1}{b^2}$

• A. $\frac{2+c}{c}$
• B. $\frac{4-c}{c}$
• C. $\frac{2-c}{c}$
• D. $\frac{1-c}{c}$

Answer 1

Answer: (C)

$\frac{2-c}{c}$

$letassumex=\alpha$

We have, $\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{{\sec }^{4}x}+\frac{1}{{cosec}^{4}x}$

$={\cos }^{4}x+{\sin }^{4}x$

$=({\sin }^{2}x{)}^2+({\cos }^{2}x{)}^2+2{\sin }^{2}x{\cos }^{2}x-2{\sin }^{2}x{\cos }^{2}x$

$=({\sin }^{2}x+{\cos }^{2}x{)}^2-2{\sin }^{2}x{\cos }^{2}x$

$=1-2{\sin }^{2}x{\cos }^{2}x$

$=1+2(\frac{1}{c}-1)=\frac{2-c}{c}$