Question

If ${\tan }^2{2}^{\circ }+{\tan }^2{4}^{\circ }+\cdots +{\tan }^2{88}^{\circ }=a,$ then the value of $\sum _{\theta =1^{\circ }}^{{89}^{\circ }}\frac{{\sin }^4\theta +{\cos }^4\theta }{{\sin }^2\theta {\cos }^2\theta }$ is

• A. $2(89+4a)$
• B. $2(89-2a)$
• C. $2(89+2a)$
• D. $2(89-4a)$

Answer 1

The correct option is A $2(89+4a)$

$\frac{{\sin }^4\theta +{\cos }^4\theta }{{\sin }^2\theta {\cos }^2\theta }=\frac{{({\sin }^2\theta -{\cos }^2\theta )}^2+2{\sin }^2\theta {\cos }^2\theta }{{\sin }^2\theta {\cos }^2\theta }$
$\Rightarrow \frac{{\sin }^4\theta +{\cos }^4\theta }{{\sin }^2\theta {\cos }^2\theta }=(\tan \theta -\cot \theta {)}^2+2$
Let $\sum _{\theta =1^{\circ }}^{{89}^{\circ }}({(\text{tan}\theta -\text{cot}\theta )}^2+2)=S$
Since $\tan \theta -\cot \theta =\frac{\sin \theta }{\cos \theta }-\frac{\cos \theta }{\sin \theta }$
$=\frac{{\sin }^2\theta -{\cos }^2\theta }{\sin \theta \cos \theta }$
$=-2\frac{\cos 2\theta }{\sin 2\theta }=-2\cot 2\theta$
$\therefore S=2\left(89\right)+4\sum _{\theta =1^{\circ }}^{{89}^{\circ }}\left({\text{cot}}^{2}2\theta \right)$

Now, ${\cot }^2{2}^{\circ }+{\cot }^2{4}^{\circ }+\cdots +{\cot }^2{178}^{\circ }$
$=2({\text{cot}}^2{2}^{\circ }+{\text{cot}}^2{4}^{\circ }+\cdots +{\text{cot}}^2{88}^{\circ })$
$=2({\text{tan}}^2{2}^{\circ }+{\text{tan}}^2{4}^{\circ }+\cdots +{\text{tan}}^2{88}^{\circ })=2a$

$\therefore S=2(89+4a)$