Question

If $sin\theta +cosec\theta =2$, the value of $si{n}^{10}\theta +cose{c}^{10}\theta$ is :

• A. $2$
• B. $2^{10}$
• C. $2^9$
• D. $10$

Answer 1

Answer: (A)

Given that, $sin\theta +cosec\theta =2...(i)$
On squaring both sides, we get
$si{n}^2\theta +cose{c}^2\theta +2=4$
$\Rightarrow si{n}^2\theta +cose{c}^2\theta ...(ii)$
Again squaring Eq. $(ii)$, we get
$si{n}^4\theta +cose{c}^4\theta =2...(iii)$
Again cubing Eq. $(ii)$, we get
$(si{n}^2\theta +cose{c}^2\theta {)}^3=2^3$
$\Rightarrow si{n}^6\theta +cose{c}^2\theta +3si{n}^2\theta cose{c}^2\theta$
$(si{n}^2\theta +cose{c}^2\theta )=8$
$\Rightarrow sin{6}^{\theta }+cose{c}^6\theta +3\cdot 2=8$
$\Rightarrow si{n}^6\theta +cose{c}^6\theta =2...(iv)$
On multiplying Eqs. $(iv)$ and $(iii)$, we get
$(si{n}^4\theta +cose{c}^4\theta )(si{n}^6\theta +cose{c}^6\theta )=4$
$\Rightarrow si{n}^{10}\theta +si{n}^4\theta cose{c}^6\theta +cose{c}^4\theta si{n}^6\theta +cose{c}^{10}\theta =4$
$\Rightarrow si{n}^{10}\theta +si{n}^4\theta cose{c}^4\theta (si{n}^2\theta +cose{c}^2\theta )+cose{c}^{10}\theta =4$
$\Rightarrow si{n}^{10}\theta +cose{c}^{10}\theta =4-2=2$