If 15sin^4α+10cos^4α=6, for some α∈R, then the value of 27sec^6α+8cosec^6α is equal to :

Correct answer: (D) 250. 15sin^4α+10cos^4α=6 15sin^4α+10cos^4α=6sin2α+cos2α^2 3sin2α-2cos2α^2=0 ∴ 3sin^2α-2cos^2α=0 tan^2α=2/3 & cot^2α=3/2 ⇒27sec^6α+8cosec^6α =27sec2α^3+8cosec2α^3 =271+tan2α^3+81+cot2α^3 =271+23^3+81+32^3 =250

Q48 / 90 · JEE Main 2021 (18 Mar Shift 2) · Mathematics

If $15 \sin^{4} α + 10 \cos^{4} α = 6,$ for some $α \in R,$ then the value of $27 \sec^{6} α + 8 {cosec}^{6} α$ is equal to :

A
$350$
B
$500$
C
$400$
Correct answer
D
$250$

Solution

$15 \sin^{4} α + 10 \cos^{4} α = 6$

$15 \sin^{4} α + 10 \cos^{4} α = 6 {(\sin^{2} α + \cos^{2} α)}^{2}$

${(3 \sin^{2} α - 2 \cos^{2} α)}^{2} = 0$

$∴ 3 \sin^{2} α - 2 \cos^{2} α = 0$

$\tan^{2} α = \frac{2}{3} & \cot^{2} α = \frac{3}{2}$

$\Rightarrow 27 \sec^{6} α + 8 {cosec}^{6} α$

$= 27 {(\sec^{2} α)}^{3} + 8 {({cosec}^{2} α)}^{3}$

$= 27 {(1 + \tan^{2} α)}^{3} + 8 {(1 + \cot^{2} α)}^{3}$

$= 27 {(1 + \frac{2}{3})}^{3} + 8 {(1 + \frac{3}{2})}^{3}$

$= 250$

Hence, the correct answer is option: 'D'.

If 15sin4α+10cos4α=6, for some α∈R, then the value of 27sec6α+8cosec6α is equal to :

Solution

Relevant Questions

If $15 \sin^{4} α + 10 \cos^{4} α = 6,$ for some $α \in R,$ then the value of $27 \sec^{6} α + 8 {cosec}^{6} α$ is equal to :