Question

If $ta{n}^4\theta +co{t}^4\theta$, where $\theta$ is an acute angle, then the value of $si{n}^4\theta +co{s}^4\theta$ is

• A. $\frac{43}{47}$
• B. $\frac{1}{\sqrt{47}}$
• C. $\frac{2}{\sqrt{47}}$
• D. $\frac{7}{9}$

Answer 1

The correct option is D $\frac{7}{9}$

$(ta{n}^2\theta {)}^2+(co{t}^2\theta {)}^2=47$

$\Rightarrow (ta{n}^2\theta +co{t}^2\theta {)}^2-2ta{n}^2\theta co{t}^2\theta =47$
$\Rightarrow (ta{n}^2\theta +co{t}^2\theta {)}^2=2+4=49$

$\Rightarrow ta{n}^2\theta +co{t}^2\theta =7$

$\Rightarrow (tan\theta +cot\theta {)}^2-2tan\theta cot\theta =7$

$\Rightarrow (tan\theta +cot\theta {)}^2=9$

$\Rightarrow {(\frac{sin\theta }{cos\theta }+\frac{cos\theta }{sin\theta })}^2=9$

$\Rightarrow {\left(\frac{si{n}^2\theta +co{s}^2\theta }{sin\theta cos\theta }\right)}^2=9$

$\Rightarrow \frac{1}{si{n}^2\theta co{s}^2\theta }=9$

$\Rightarrow si{n}^2\theta co{s}^2\theta =\frac{1}{9}$

$\therefore si{n}^2\theta co{s}^2\theta =(si{n}^2\theta +co{s}^2\theta {)}^2-2si{n}^2\theta co{s}^2\theta$

$=1-2\times \frac{1}{9}=\frac{7}{9}$