JEE Mains · Maths · STD 11 - 3. trignometrical ratios,functions and identities
If \(10 \sin ^4 \theta+15 \cos ^4 \theta=6\), then the value of \(\frac{27 \operatorname{cosec}^6 \theta+8 \sec ^6 \theta}{16 \sec ^8 \theta}\) is:
- A \(\frac{2}{5}\)
- B \(\frac{3}{4}\)
- C \(\frac{3}{5}\)
- D \(\frac{1}{5}\)
Answer & Solution
Correct Answer
(A) \(\frac{2}{5}\)
Step-by-step Solution
Detailed explanation
\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 \\…
See the Complete Solution
Get step-by-step explanations for this and 2.5 Lakh+ more JEE, NEET & CET questions.
- Unlock all solutions
- Practice the full chapter
- Track accuracy across PYQs
4.8 rated on Google Play · 14,000+ reviews
More questions from Maths
- Let \(\alpha, \beta\) be roots of \(x^2+\sqrt{2} x-8=0\). If \(\mathrm{U}_{\mathrm{n}}=\alpha^{\mathrm{n}}+\beta^{\mathrm{n}}\), then \(\frac{\mathrm{U}_{10}+\sqrt{2} \mathrm{U}_9}{2 \mathrm{U}_8}\) is equal to ............JEE Mains 2024 Medium
- \(PQR\) triangular park with \(PQ = PR = 200\ m.A\) TV tower stands at the mid-point of \(QR\). If the angles of elevation of the top of the tower at \(P, Q\) and \(R\) are respectively \(45^o , 30^o \) and \(30^o \), then the height of the tower \((in \,m)\) is:JEE Mains 2018 Hard
- Let the direction cosines of two lines satisfy the equations: \( 4l+m-n=0 \) and \( 2mn+10nl+3lm=0 \). Then the cosine of the acute angle between these lines is:JEE Mains 2026 Easy
- Considering only the principal values of the inverse trigonometric functions, the domain of the function \(f(x)=\cos ^{-1}\left(\frac{x^{2}-4 x+2}{x^{2}+3}\right)\) is.JEE Mains 2022 Medium
- If the system of linear equations \(2x + 2y + 3z = a\) ; \(3x - y + 5z = b\) ; \(x - 3y + 2z = c\) Where \(a, b, c\) are non zero real numbers, has more than one solution, thenJEE Mains 2019 Hard
- Let the latus ractum of the parabola \(y ^{2}=4 x\) be the common chord to the circles \(C _{1}\) and \(C _{2}\) each of them having radius \(2 \sqrt{5}\). Then, the distance between the centres of the circles \(C _{1}\) and \(C _{2}\) isJEE Mains 2020 Medium
More PYQs from JEE Mains
- If \(f: \mathbf{N} \rightarrow \mathbf{Z}\) is defined by
\(f(n) = \begin{vmatrix} n & -1 & -5 \\ -2n^2 & 3(2k+1) & 2k+1 \\ -3n^3 & 3k(2k+1) & 3k(k+2)+1 \end{vmatrix}\), \(k \in \mathbf{N}\),
and \(\sum_{n=1}^{k} f(n) = 98\), then \(k\) is equal to :JEE Mains 2026 Hard - If the system of linear equations \(x+ ay+z\,= 3\) ; \(x + 2y+ 2z\, = 6\) ; \(x+5y+ 3z\, = b\) has no solution, thenJEE Mains 2018 Hard
- If \((x, y, z)\) be an arbitrary point lying on a plane \(P\) which passes through the point \((42,0,0) , (0,42,0)\) and \((0,0,42),\) then the value of expression \(3+\frac{x-11}{(y-19)^{2}(z-12)^{2}}+\frac{y-19}{(x-11)^{2}(z-12)^{2}}\)\( +\frac{z-12}{(x-11)^{2}(y-19)^{2}}-\frac{x+y+z}{14(x-11)(y-19)(z-12)} \)JEE Mains 2021 Hard
- Let \(y(x)\) be the solution of the differential equation \(\left( {xlogx} \right)\frac{{dy}}{{dx}} + y = 2xlogx,\left( {x \ge 1} \right)\) . Then \(y(e) \) is equal to : \([y(1)=0]\)JEE Mains 2015 Hard
- Let \(f: \mathbb{R} \rightarrow \mathbb{R}\) be a function defined by
\(f(x)=(2+3 a) x^2+\left(\frac{a+2}{a-1}\right) x+b, a \neq 1 . \text { If }\)
\(f(x+\mathrm{y})=f(x)+f(\mathrm{y})+1-\frac{2}{7} x \mathrm{y}\), then the value of \(28 \sum_{i=1}^5|f(i)|\) isJEE Mains 2025 Medium - If the function \(f:(-\infty,-1] \rightarrow(a, b]\) defined by \(f(x)=e^{x^3-3 x+1}\) is one-one and onto, then the distance of the point \(\mathrm{P}(2 \mathrm{~b}+4, \mathrm{a}+2)\) from the line \(x+e^{-3} y=4\) is :JEE Mains 2024 Hard