Question

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

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

Answer 1

Answer: (A)

Given that, $sin \,θ + cosec\,θ = 2 \quad ...(i)$
On squaring both sides, we get
$ sin^2\,θ + cosec^2 \,θ +2 = 4$
$ ⇒ sin^2\,θ + cosec^2\,θ \quad...(ii)$
Again squaring Eq. $(ii)$, we get
$sin^4\,θ + cosec^4 \,θ = 2\quad ...(iii)$
Again cubing Eq. $(ii)$, we get
$(sin^2\,θ +cosec^2 \,θ)^3 = 2^3$
$ ⇒ sin^6 \,θ + cosec^2\,θ + 3\,sin^2 \,θ \,cosec^2\,θ$
$(sin^2 \,θ +cosec^2\,θ) = 8$
$⇒ sin6^ θ + cosec^6 \,θ + 3 \cdot 2 = 8 $
$⇒ sin^6\,θ + cosec^6\,θ = 2 \quad ...(iv)$
On multiplying Eqs. $(iv)$ and $(iii)$, we get
$(sin^4 \,θ + cosec^4\,θ)(sin^6\,θ + cosec^6\,θ) = 4$
$ ⇒ sin^{10} \,θ + sin^4 \,θ \, cosec^6 \,θ + cosec^4 \,θ \,sin^6 \,θ+ cosec^{10}\,θ = 4$
$ ⇒ sin^{10} θ + sin^4\, θ \,cosec^4\, θ (sin^2 \, θ + cosec^2\, θ) + cosec^{10}\, θ = 4$
$ ⇒ sin^{10} \, θ + cosec^{10}\, θ = 4 - 2 = 2$