JEE Main 2025 4 April Morning Shift
Exam JEE Main
Year 2025
Date 4 April
Shift Morning Shift
Subject Mathematics
Chapter Trigonometric Equations JEE Main 2025 · 4 April · Morning Shift · Mathematics · Trigonometric Equations
If 10 ^4 θ+15 ^4 θ=6 , then the value of 27 cosec^6 θ+8 ^6 θ 16 ^8...
If
10 sin 4 θ + 15 cos 4 θ = 6 10 \sin ^4 \theta+15 \cos ^4 \theta=6 10 sin 4 θ + 15 cos 4 θ = 6 , then the value of
27 cosec 6 θ + 8 sec 6 θ 16 sec 8 θ \frac{27 \operatorname{cosec}^6 \theta+8 \sec ^6 \theta}{16 \sec ^8 \theta} 16 s e c 8 θ 27 cosec 6 θ + 8 s e c 6 θ is:
Options A 2 5 \frac{2}{5} 5 2 B 3 4 \frac{3}{4} 4 3 C 3 5 \frac{3}{5} 5 3 D 1 5 \frac{1}{5} 5 1 Correct answer A. 2 5 \frac{2}{5} 5 2
Step-by-step solution
10 ( sin 2 θ ) 2 + 15 ( 1 − sin 2 θ ) 2 = 6 Let sin 2 θ = t ⇒ 10 t 2 + 15 ( 1 − t ) 2 = 16 10 t 2 + 15 − 30 t + 15 t 2 = 6 25 t 2 − 30 t + 9 = 0 ( 5 t − 3 ) 2 = 0 sin 2 θ = 3 5 and cos 2 θ = 2 5 27 × 125 27 + 8 + 125 8 16 ( 5 2 ) 4 = 250 125 × 5 = 2 5 \begin{aligned} & 10\left(\sin ^2 \theta\right)^2+15\left(1-\sin ^2 \theta\right)^2=6 \\ & \text{ Let } \sin ^2 \theta=\mathrm{t} \Rightarrow 10 \mathrm{t}^2+15(1-\mathrm{t})^2=16 \\ & 10 \mathrm{t}^2+15-30 \mathrm{t}+15 \mathrm{t}^2=6 \\ & 25 \mathrm{t}^2-30 \mathrm{t}+9=0 \\ & (5 \mathrm{t}-3)^2=0 \\ & \sin ^2 \theta=\frac{3}{5} \text{\text{ and } } \cos ^2 \theta=\frac{2}{5} \\ & \frac{27 \times \frac{125}{27}+8+\frac{125}{8}}{16\left(\frac{5}{2}\right)^4}=\frac{250}{125 \times 5}=\frac{2}{5}\end{aligned} 10 ( sin 2 θ ) 2 + 15 ( 1 − sin 2 θ ) 2 = 6 Let sin 2 θ = t ⇒ 10 t 2 + 15 ( 1 − t ) 2 = 16 10 t 2 + 15 − 30 t + 15 t 2 = 6 25 t 2 − 30 t + 9 = 0 ( 5 t − 3 ) 2 = 0 sin 2 θ = 5 3 and cos 2 θ = 5 2 16 ( 2 5 ) 4 27 × 27 125 + 8 + 8 125 = 125 × 5 250 = 5 2
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